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A TEXTBOOK 


ON 

THE GAS ENGINE 


International Correspondence Schools 

*' SCRANTON, PA. 


ANSWERS TO QUESTIONS 


1943 


SCRANTON 

INTERNATIONAL TEXTBOOK COMPANY 
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CONTENTS. 


Arithmetic - . . - 

Mensuration and Use of Let¬ 
ters in Formulas, 
Elementary Algebra and Trig¬ 
onometric Functions, - 
Logarithms . . . - 

Elementary Mechanics - 
Pneumatics, Gas, and Petro¬ 
leum, . 

Heat,. 

Gas, Gasoline, and Oil 
Engines, . . . - 


Answers to Questions 1 

Answers to Questions 93 

• 

Answers to Questions 109 
Answers to Questions 131 
Answers to Questions 139 

Answers to Questions 161 
Answers to Questions 169 

Answers to Questions 183 






NOTICE. 


There is a break in the continuity of the page numbers and 
the question numbers between the answers to questions rela¬ 
ting to Arithmetic and those relating to the section on 
Mensuration and Use of Letters in Formulas, and in the 
continuity of the question numbers and figure numbers 
between the answers to questions relating to the section on 
Logarithms and those relating to the section on Elementary 
Mechanics, and between those relating to the section on 
Elementary Mechanics and those relating to the section on 
Pneumatics, Gas, and Petroleum. This does not affect the 
subject matter, which is published in full, and in regular 
order. 







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ARITHMETIC 


(SECTIONS 1-3.) 
(QUESTIONS 1-75.) 


(1) See Art. 1. 

(2) See Art. 3. 

( 3 ) See Arts. 5 and 6 . 

(4) See Arts. 10 and 11. 

( 5 ) 980 = Nine hundred eighty. 

605 = Six hundred five. 

28,284 = Twenty-eight thousand,two hundred eighty-four, 
9,006,042 = Nine million, six thousand and forty-two. 
850,317,002= Eight hundred fifty million, three hundred 
seventeen thousand and two. 

700,004 = Seven hundred thousand and four. 

(6) Seven thousand six hundred = 7,600. 

Eighty-one thousand four hundred two = 81,402. 

Five million, four thousand and seven = 5,004,007. 

One hundred and eight million, ten thousand and one = 
108,010,001. 

Eighteen million and six = 18,000,006. 

Thirty thousand and ten = 30,010. 


(7) In adding whole numbers, place the numbers to be 


added directly under each other so that 
the extreme right-hand figures will stand 
in the same column, regardless of the 
position of those at the left. Add the first 
column of figures at the extreme right, 
which equals 19 units, or 1 ten and 9 
units. We place 9 units under the units 
column, and reserve 1 ten for the column 


3290 

504 

865403 

2074 

81 

7 


8 713 5 9 Ana. 

For notice of the copyright, see page immediately following the title page 




2 


ARITHMETIC. 


of tens. 14-8 + 7 + 9 = 25 tens, or 2 hundreds and 5 
tens. Place 5 tens under the tens column, and reserve 
2 hundreds for the hundreds column. 2 + 4+5 + 2 = 13 
hundreds, or 1 thousand and 3 hundreds. Place 3 hundreds 
under the hundreds column, and reserve the 1 thousand 
for the thousands column. 1 + 2 + 5 + 3=11 thousands, 
or 1 ten-thousand and 1 thousand. Place the 1 thousand in 
the column of thousands, and reserve the 1 ten-thousand 
for the column of ten-thousands. 1 + G = 7 ten-thousands. 
Place this seven ten-thousands in the ten-thousands column. 
There is but one figure 8 in the hundreds of thousands place 
in the numbers to be added, so it is placed in the hundreds 
of thousands column of the sum. 

A simpler (though less scientific) explanation of the same 
problem is the following; 7 + 1 + 4+ 3 + 4+ 0 = 19; write 
the nine and reserve the 1. 1 + 8+ 7 + 0 + 0 + 9= 25; 

write the 5 and reserve the 2. 2 + 0 + 4 + 5 + 2=13; 

write the 3 and reserve the 1. 1 + 2 + 5 + 3=11; write 

the 1 and reserve 1. 1+6 = 7; write the 7. Bring down 

the 8 to its place in the sum. 

( 8 ) 709 

8304725 

391 

100302 

300 

909 

8407336 Ans. 

( 9 ) (a) In subtracting whole numbers, place the sub- 
trahend or smaller number under the minuend or larger 
number, so that the right-hand figures stand directly under 
each other. Begin a/ the right to subtract. We can not 
subtract 8 units from 2 units, so we take 1 ten from the 
6 tens and add it to the 2 units. As 1 ten = 10 units, we 
have 10 units + 2 units =12 units. Then, 8 units from 
12 units leaves 4 units. We took 1 ten from 6 tens, so 



ARITHMETIC. 


only 5 tens remain. 3 tens from 5 tens 50952 
leaves 2 tens. In the hundreds column we 3333 

have 3 hundreds from 9 hundreds leaves - 

G hundreds. We can not subtract 3 thou- Ans. 

sands from 0 thousands, so we take 1 ten-thousand from 
5 ten-thousands and add it to the 0 thousands. 1 ten^ 
thousand =10 thousands^ and 10 thousands -}- 0 thousands 
= 10 thousands. Subtracting, we have 3 thousands from 
10 thousands leaves 7 thousands. We took 1 ten-thousand 
from 5 ten-thousands and have 4 ten-thousands remaining. 
Since there are no ten-thousands in the subtrahend, the 
4 in the ten-thousands column in the minuend is brought 
down into the same column in the remainder, because 0 from 
4 leaves 4. 

{b) 1533 9 
10001 

5338 Ans, 

(10) (a) 709G8 {b) 100000 

32975 98735 

3 7 9 9 3 Ans. 12 6 5 Ans. 

(11) We have given the minuend or greater number 
(1,004) and the difference or remainder (49). Placing these 

1004 

in the usual form of subtraction we have - in which 

49 

the dash (-) represents the number sought. This number 

is evidently less than 1,004 by the difference 49, hence, 
1,004 — 49 = 955, the smaller number. For the sum of the 

10 0 4 larger 

two numbers we then have 9 5 5 smaller 

1 9 5 9 sum. Ans. 

Or, this problem may be solved as follows: If the greater 
of two numbers is 1,004, and the difference between them is 
49 , then it is evident that the smaller number must be 
equal to the difference between the greater number (1,004) 










4 


ARITHMETIC, 


and the difference (49); or, 1,004 — 49 = 955, the smaller 
number. Since the greater number equals 1,004 and the 
smaller number equals 955, their sum equals 1,004 + 955 
= 1,959 sum. Ans. 

(12) The numbers connected by the plus (+) sign must 
first be added. Performing these operations we have 
5962 3874 

8471 2039 

9 0 2 3 5 913 sum. 

2 3 4 5 6 sum. 

Subtracting the smaller number (5,913) from the greater 
(23,456) we have 

23456 

5913 

17 5 4 3 difference. Ans. 


( 13 ) 14 4 6 7 5 = amount willed to his son. 

2 6 3 8 0 = amount willed to his daughter. 
$71055 = amount willed to his two children. 


$ 


12 5 0 0 0 = amount willed to his wife and two 
children. 

710 5 5 = amount willed to his two children. 
$53945 = amount willed to his wife. Ans. 


(14) In the multiplication of whole numbers, place the^ 
multiplier under the multiplicand, and multiply each term 
of the multiplicand by each term of the multiplier, writing 
the right-hand figure of each product obtained under the 
term of the multiplier which produces it. 


7x7 units = 49 units, or 4 tens and 9 
units. We write the 9 units and reserve 
the 4 tens. 7 times 8 tens = 56 tens; 
56 tens + 4 tens reserved = 60 tens or 
6 hundreds and 0 tens. Write the 0 
tens and reserve the 6 hundreds. 7x3 hundreds = 21 hun¬ 
dreds; 21 + 6 hundreds reserved = 27 hundreds, or 2 thou¬ 
sands and 7 hundreds. Write the 7 hundreds and reserve 


{a) 

526387 

_7 

3 6 8 4 7 0 9 Ans. 








ARITHMETIC. 


5 


the 2 thousands. 7x6 thousands = 42 thousands; 42 
-j- 2 thousands reserved = 44 thousands or 4 ten-thousands 
and 4 thousands. Write the 4 thousands and reserve the 
4 ten-thousands. 7x2 ten-thousands = 14 ten-thousands; 
14 4- 4 ten-thousands reserved = 18 ten-thousands, or 
1 hundred-thousand and 8 ten-thousands. Write the 8 ten- 
thousands and reserve the 1 hundred-thousand. 7x5 hun¬ 
dred-thousands = 35 hundred-thousands; 35-f-1 hundred- 
thousand reserved = 36 hundred-thousands. Since there 
are no more figures in the multiplicand to be multiplied, 
we write the 36 hundred-thousands in the product. This 
completes the multiplication. 

A simpler (though less scientific) explanation of the same 
problem is the following: 

7 times 7 = 49;. write the 9 and reserve the 4. 7 times 

8 = 56; 56 -j- 4 reserved = 60; write the 0 and reserve the 6. 
7 times 3 = 21; 21 + 6 reserved = 27; write the 7 and re¬ 
serve the 2. 7 X 6 = 42; 42 + 2 reserved =• 44; write the 

4 and reserve 4. 7 X 2 = 14; 14 + 4 reserved = 18; write 

the 8 and reserve the 1. 7 X 5 = 35; 35 + 1 reserved = 36; 

write the 36. 

In this case the multiplier is 17 
units^ or 1 ten and 7 units^ so that 
the product is obtained by adding 
two partial products, namely, 7 X 
700,298 and 10 X 700,298. The 
actual operation is performed as 
follows: 

7 times 8 = 56; vTite the 6 and reserve the 5. 7 times 9 = 

63; 63 + 5 reserved =68; write the 8 and reserve the 6. 
7 times 2 = 14; 14+6 reserved = 20; write the 0 and re¬ 
serve the 2. 7 times 0 = 0; 0 + 2 reserved = 2; write the 2. 

7 times 0 = 0; 0 + 0 reserved = 0; write the 0. 7 times 7 = 

49; 49 + 0 reserved = 49; write the 49. 

To multiply by the 1 ten we say 1 times 700298 = 700298, 
and write 700298 under the first partial product, as shown, 
with the right-hand figure 8 under the multiplier 1. Add the 
two partial products: their sum equals the entire product. 


{b) 


700298 

_ 

4902086 

700298 

11905066 Ans. 




6 


ARITHMETIC. 


(c) 217 

103 
651 
2170 
22351 
67 

156457 

134106 

1497517 


Multiply any two of the numbers together 
and multiply their product by the third 
number. 


Ans. 


(15) If your watch ticks every second, then to find how 
many times it ticks in one week it is necessary to find the 
number of seconds in 1 week. 

6 0 seconds = 1 minute. 

6 0 rninutes = 1 hour. 

3 6 0 C seconds = 1 hour. 

2 4 hours = 1 day. 

14400 

7200 

8 6 4 0 0 seconds = 1 day. 

7 days = 1 week. 

6 0 4 8 0 0 seconds in 1 week or the number of times that 
Ans. your watch ticks in 1 week. 

( 16 ) If a monthly publication contains 24 pages, a yearly 

2 4 volume will contain 12x24 or 288 pages, since 
12 there are 12 months in one year; and eight 

2 3 g yearly volumes will contain 8x288, or 2,304 

8 pages. 

2 3 0 4 Ans. 

( 17 ) If an engine and boiler are worth $3,246, and the 
building is worth 3 times as much, plus $1,200, then the 
building is worth 

$3246 

3 

9738 
plus 12 0 0 

$10 9 3 8 = value of building. 













ARITHMETIC. 


7 


If the tools are worth twice as much as the building, plus 
$1,875, then the tools are worth * 


$10938 

2 

21876 
plus 18 7 5 

$23751= 

Value of building =$10938 
Value of tools = 2 3 7 51 

$346 89 = 

Value of engine and 

boiler = $ 3 2 4 6 
Value of building 

and tools = 3 4 6 8 9 

$37 9 35 = 


value of tools. 


value of the building 
and tools, (a) Ans. 


value of the whole 
plant, {b) Ans. 


( 18 ) {a) (72 X 48 X 28 X 5) -J- (96 X 15 X 7 X 6). 
Placing the numerator over the denominator the problem 
becomes 

72 X 48 X 28 X 5 _ 

96 X 15 X 7 X 6 ~ 


The 5 in the dividend 15 in the divisor are both divis¬ 
ible by 5, since 5 divided by 5 equals 1, and 15 divided by 
5 equals 3. Cross off the 5 and write the 1 over it; also cross 
off the 15 and write the 3 under it. Thus, 

1 

72 X 48 X 28 X ^ _ 

96 X X 7 X 6 
3 


The 5 and 15 are not to be considered any longer, and, in 
fact, may be erased entirely and the 1 and 3 placed in their 
stead, and treated as if the 5 and 15 never existed. Thus, 

72 X 48 X 28 X 1 


96 X 3 X 7 X 6 









8 


ARITHMETIC. 


72 in the dividend and 9G in the divisor are divisible by 12, 
since 72 divided by 12 equals 6, and 00 divided by 12 equals 
8. Cross offCciQ 72 and write the 6 over it; also, cross off 
the 96 and write the 8 under it. Thus, 

6 

X 48 X 28 X 1 ^ 

X 3*x 7 X 6 

8 

The 72 and 96 are not to be considered any longer, and, 
in fact, may be erased entirely and the 6 and 8 placed in 
their stead, and treated as if the 72 and 96 never existed. 
Thus, 

6 X 48 X 28 X 1 _ 

8X3X7X6 “ 

Again, 28 in the dividend and 7 in the divisor are divisible 
by 7, since 28 divided by 7 equals 4, and 7 divided by 7 
equals 1. Cross offCa^ 28 and write the 4 over it; also, cross 
off the 7 and write the 1 under it. Thus, 

4 

6 X 48 X X 1 ^ 

8 X 3 X y X 6 
1 

The 28 and 7 are not to be considered any longer, and, in 
fact, may be erased entirely and the 4 and 1 placed in their 
stead, and treated as if the 28 and 7 never existed. Thus, 
6X48X4X1 _ 

8X3X1X6 ~ 

Again, 48 in the dividend and 6 in the divisor are divisible 
by 6, since 48 divided by 6 equals 8, and 6 divided by 6 equals 
1. Cross off 48 and write the 8 over it; also, cross off 
the 6 and write the 1 under it. Thus, 

8 

6x^^x4xl _ 

8 X 3 X 1 X^ 

1 

The 48 and 6 are not to be considered any longer, and, in 
fact, may be erased entirely and the 8 and 1 placed in their 
stead, and treated as if the 48 and 6 never existed. Thus, 







ARITHMETIC. 


9 


6 X 8 X 4 X 1 
8 X 3 X 1X 1“ 

Again, 6 in the dividend and 3 in the divisor are divisible 
by 3, since 6 divided by 3 equals 2, and 3 divided by 3 equals 
1 . Cross off the 6 and write the 2 over it; also, cross off the 
3 and write the 1 under it. Thus, 

2 

^X 8 X 4 X 1 _ 

8x^xlxl“ 

1 

The 6 and 3 are not to be considered any longer, and, in 
fact, may be erased entirely and the 2 and 1 placed in their 
stead, and treated as if the 6 and 3 never existed. Thus, 


2 X 8 X 4 X 1 _ 

8X1X1X1“ 

Canceling the 8 in the dividend and the 8 in the divisor, 
the result is 

1 

2x^x4xl _ 2xlx4xl 
gxlxlxl ixlxlxl* 

1 


Since there are no two remaining numbers (one in the 
dividend and one in the divisor) divisible by any number ex¬ 
cept 1, without a remainder, it is impossible to cancel further. 

Multiply all the uncanceled numbers in the dividend 
together, and divide their product by the product of all 
the uncanceled numbers in the divisor. The result will be the 
qtiotient. The product of all the tincanceled numbers in 
the dividend equals 2XlX4xl = 8; the product of all the 
uncanceled numbers in the divisor equals lx 1X1X1= 1. 


Hence, 


Or, 


2X1X4X1 8 

IXlXlXl 1 


F.x^ x f xLs 8. Ans. 
X X ST X ^ 1 


U.O. IV. 









10 


ARITHMETIC. 


{/>) (80 X 60 X 50 X 16 X 14) -i- (70 X 50 X 24 X 20). 

Placing the numerator over the denominator, the problem 
becomes 

80 X 60 X 50 X 16 X 14 _ , 

70 X 50 X 24 X 20 

The 50 in the dividend and 70 in the divisor are both divis¬ 
ible hy 10^ since 50 divided by 10 equals 5, and 70 divided 
by 10 equals 7. Cross off the 50 and write the 5 over it; 
also, cross off the 70 and write the 7 under it. Thus, 

5 

80 X 60 X X 16 X 14 __ 

X 50 X 24 X 20 

7 

The 50 and 70 are not to be considered any longer, and, 
in fact, may be erased entirely and' the 5 and 7 placed in 
their stead, and treated as if the 50 and 70 never existed. 
Thus, 

80 X 60 X 5 X 16 X 14 _ 

7 X 50 X 24 X 20 “ 

Also, 80 in the dividend and 20 in the divisor are divisible 
by 20, since 80 divided by 20 equals 4, and 20 divided by 20 
equals 1. Cross off the 80 and write the 4 over it; also, 
cross off the 20 and write the 1 under it. Thus, 

4 

X 60 X 5 X 16 X 14 _ 

7 X 50 X 24 X 

1 

The 80 and 20 are not to be considered any longer, and, 
in fact, may be erased entirely and the 4 and 1 placed in 
their stead, and treated as if the 80 and 20 never existed. 
Thus, 

4 X 60 X 5 X 16 X 14 _ 

7 X 50 X 24 X 1 

Again, 16 in the dividend and 24 in the divisor are divisible 
by 8, since 16 divided by 8 equals 2, and 24 divided by 8 
equals 3. Cross off the 16 and write the 2 over it; also cross 
off the 24 and write the 3 under it. Thus, 







ARITHMETIC. 


11 


2 

4 X 60 X 5 X X 14 
7 X 50 X 524 X 1 

3 

The 16 and 24 are not to be considered any longer, and, 
in fact, may be erased entirely and the 2 and 3 placed in 
their stead, and treated as if the 16 and 24 never existed. 
Thus, 

4X60X5X2X14 
7X50X3X1 “ 

Again, 60 in the dividend and 50 in the divisor are divis¬ 
ible by 10, since 60 divided by 10 equals 6, and 50 divided by 
10 equals 5. Cross <?^the 60 and write the 6 over it; also, 
cross off the 50 and write the 5 under it. Thus, 

6 

4x^0x5x2xl4 _ 

7x ^(5x3x1 “ 

6 

The 60 and 50 are not to be considered any longer, and, in 
fact, may be erased entirely and the 6 and 5 placed in their 
stead, and treated as if the 60 and 50 never existed. Thus, 

4X6X5X2X14_ 

7 X 5 X 3 X 1 

The 14 in the dividend and 7 in the divisor are divisible by 
7, since 14 divided by 7 equals 2, and 7 divided by 7 equals 1. 
Cross off the 14 and write the 2 over it; also, cross off the 7 
and write the 1 under it. Thus, 

2 

4x6x5x2x;^ _ 

;r X 5 X 3 X 1 

1 

The 14 and 7 are not to be considered any longer, and, in 
fact, may be erased entirely and the 3 and 1 placed in their 
stead, and treated as if the 14 and 7 never existed. Thus, 

4X6X5X2X2_ 


1 X 5 X 3 X 1 








12 


ARITHMETIC. 


The 5 in the dividend and 5 in the divisor are divis¬ 
ible by 5, since 5 divided by 5 equals 1. Cross off the 5 
of the dividend and write the 1 over it; also, cross off the 5 
of the divisor and write the 1 under it. Thus, 

1 

4x6x ^X2X2^ 

1 X ^ X 3 X 1 

1 


The 5 in the dividend and 5 in the divisor are not to be 
considered any longer, and, in fact, may be erased entirely 
and 1 and 1 placed in their stead, and treated as if the 5 and 
5 never existed. Thus, 

4X6X1X2X2 _ 

1X1X3X1~ 


The 6 in the dividend and 3 in the divisor are divisible by 
3, since 6 divided by 3 equals 2, and 3 divided by 3 equals 1. 
Cross off the 6 and place 2 over it; also, cross off the 3 and 
place 1 under it. Thus, 

2 

4x^xlx2x2 _ 

ixlx^xl “ 

1 


The 6 and 3 are not to be considered any longer, and, in 
fact, may be erased entirely and 2 and 1 placed in their 
stead, and treated as if the 6 and 3 never existed. Thus, 


4x2xlx2x2 _32_ 
ixlxlxl “1“ 


Ans. 


2 1 

4 ^ ^ 2 2 

Hnnor X W X ^0 X X _ 4 X 2 X1 X 2 X 2 32 
’ yM^0x#x^p “ ixlxlxl ”T“ 
y ^ ^ 1 ^ 

111 


( 19 ) 28 acres of land at $133 an acre would cost 

28 X $133 = $3,724. 

28 

1064 

266 


$3724 










ARITHMETIC. 


13 


If a mechanic earns 11,500 a year and his expenses are 
$968 per year, then he would save $1500—1968, or $532 
per year. 9 6 8 

$532 

If he saves $532 in 1 year, to save $3,724 it would take as 
many years as $532 is contained times in $3,724, or 7 years. 

532)3724(7 years. Ans. 

3724 


(20) If the freight train ran 365 miles in one week, and 

3 times as far lacking 246 miles the next week, then it ran 
(3 X 365 miles) — 246 miles, or 849 miles the second week. 
Thus, 3 6 5 

3 

ToTs 

246 

difference 8 49 miles. Ans. 

( 21 ) The distance from Philadelphia to Pittsburg is 354 
miles. Since there are 5,280 feet in one mile, in 354 mile? 
there are 354 X 5,280 feet, or 1,869,120 feet. If the driving 
wheel of the locomotive is 16 feet in circumference, then in 
going from Philadelphia to Pittsburg, a distance of 1,869,- 
120 feet, it will make 1,869,120 -f-16, or 116,820 revolutions. 

16)1869120(116820 rev. Ans. 








14 


ARITHMETIC. 


(22) («) 576)589824(102 4 Ans 

576 
1382 
1152 

2304 

2304 


(^) 43911 ) 369730620 (8420 Ans. 

351288 
184426 
175644 

87822 

87822 

0 

(c) 505 ) 2527525 ( 5005 Ans. 

2525 

2525 

2525 


(d) 1234)4961794302(4020903 Am 

4936 

2579 

2468 

11143 

11106 

3702 

3702 

(23) The harness evidently cost the difference between 
$444 and the amount which he paid for the horse and wagon. 

Since $264 + $153 = $417, the amount paid for the horse 
and wagon, $444 — $417 = $27, the cost of the harness. 

$264 $444 

153 417 

$ 2 7 Ans. 


$417 














ARITHMETIC. 


15 


(24) (a) 1024 

576 

6144 

7168 

6120 

689824 Ans. 

(d) 50 0 5 

505 

25025 

250250 

^5 2 7 5.2 6 Ans. 

(c) 43 911 

8420 

878220 

175644 

351288 

369730620 Ans. 


(25) Since there are 12 months in a year, the number of 

days the man works is 25 X 12 = 300 days. As he works 10 
hours each day, the number of hours that he works in one 
year is 300 x 10 = 3,000 hours. Hence, he receives for his 
work 3,000 X 30 = 90,000 cents, or 90,000 100 = $900. Ans. 

(26) See Art. 71. 

(27) See Art. 77. 

(28) See Art. 73- 

(29) See Art. 73. 

(30) See Art. 75. 


13 

(31) — is an improper fraction, since its numerator 13 

8 

is greater than its denominator 8. 


1 ..3 ,.4 








16 


ARITHMETIC. 


(33) To reduce a fraction to its lowest terms means to 
change its form without changing its value. In order to do 
this, we must divide both numerator and denominator by 
the same number until we can no longer find any num¬ 
ber (except 1 ) which will divide both of these terms without 
a remainder. 


To reduce the fraction — to its lowest terms we divide 

o 

both numerator and denominator by 4, and obtain as a 

1 4—41. 4-^-4 

result the fraction —. Thus, - * = .t ; similarly, = 

Z o -T- 4 /y lb -T- 4 


4’ 32-^4 “8-^2 4^64-^8“8-^4 2* 


(34) When the denominator of any number is not 
expressed, it is understood to be 1 , so that ^ is the same as 

6 -h 1, or 6 . To reduce j to an improper fraction whose 

denominator is 4, we must multiply both numerator and 
denominator by some number which will make the denomi¬ 
nator of 6 equal to 4. Since this denominator is 1, by mul¬ 
tiplying both terms of ^ by 4 we shall have ^ ^ f 
t' J ^ I J 1X44’ 

which has the sa7ne value as 6 , but has a different form. Ans. 

(35) In order to reduce a mixed number to an improper 
fraction, we must multiply the whole nmnber by the denom¬ 
inator of the fractio7t and add the numerator of the fraction 
to that product. This result is the numerator of the improper 
fraction., of which the denominator is the denominator of the 
fractional part of the mixed number. 


7 7 

7— means the same as 7 5 -* 

8 8 


In 1 there are—, hence in 
8 


.U 8 56 56 , u 7 . ^ 

7 there are 7 X 5 - = - 5 -; - 3 - plus the — of the mixed number 

000 8 

= ^ ^ which is the required improper fraction. 

8 8 8 


(13X16)+ 5 213 3 

^"*10 16 16 ’ 


(10 X 4) + 3_43 
4 “4’ 




ARITHMETIC. 


11 


(36) The value of a fraction is obtained by dividing the 
numerator by the denominator. 

13 

To obtain the value of the fraction we divide the num- 

Z 

erator 13 by the denominator 2. 2 is contained in 13 six 

times, with 1 remaining. This 1 remaining is written over 

the denominator 2 , thereby making the fraction i which is 


annexed to the whole number G, and we obtain 6 — as the 

mixed number. The reason for performing this operation is 

2 13 

the following: In 1 there are — (two halves), and in — (thir- 

teen halves) there are as many units ( 1 ) as 2 is contained 

times in 13, which is 6 , and ~ (one-half) unit remaining. 

Z 


13 11 

Hence, — = 6 + ^ required mixed number. 

12 2 2 


H = 4i 

4 4 


Ans. 


69 

16 




Ans. ^ = 2. 

O 


Ans. 


64' 


Ans. 


Ans. 

67 _ 
64 


(37) In division of fractions, invert the divisor (or, in 
other words, turn it upside down) and proceed as in multi¬ 
plication. 


/ N 5 _ 35 ^16 

{a) 35 -T- - Y X y 


35 X 16 
1X5 



Ans. 


(p) 

(f) 

(d) 

m 

28)118(4^. Ans. 


V O . 

= — = —. Ans. 


9 


9 

. 3 _ 

_ 9 

1 

9 X 

1 

9 

3 

16 * 

• 3 = 

16 

* 1 " 

"" 16 

^ 3 

16 X 

3 

“48““ 

16* 

I'J' . 

■ 9 = 

17 

9 _ 

_ ^7 

X - 

17X 

1 

_ 17 

Ans. 

2 ' 


2 

* 1 ' 

" 2 

9 

■ 2X 

9 

“ 18' 


113 

7 


113 

16 

113 X 16 


1,808 _ 

452 

64 ' 

* 16 

=' 

X 

CO 

7 ‘ 

64 X 7 


448 

112 


112 


1 






18 


ARITHMETIC. 


(^) i 5 _ 4 I r= ? Before proceeding with the division, 

^ ^ 4 8 

reduce both of the mixed numbers to improper fractions. 

3_(4x8) + 3_ 




Thus 


32 + 3 35 


8 


63 35 


—. The problem is now ~ ^ before, 


invert 


63 35 63 8 63 X 8 


the divisor and multiply; -^•^■^ = X^ 35 “ 4 x 35 


504 _ 252 _ 126 _ 18 
140 “ 70 “ 35 " 5 * 


5)18(3- 
15 ' 


Ans. 


( 38 ) 


8 ^ 8 ^ 8 


l + a + 5 
8 


I-' 


Ans. 


When the denominators of the fractions to be added are 
alike^ we know that the units are divided into the same 
number of parts (in this case eighths) ; we, therefore, add the 
mimerators of the fractions to find the number of parts 


(eighths) taken or considered, thereby obtaining 
the sum. 


or 1 as 


(39) When the denominators are not alike we know that 
the units are divided into unequal parts, so before adding 
them we must find a common denominator for the denom¬ 
inators of all the fractions. Reduce the fractions to fractions 
having this common denominator, add the numerators and 
write the sum over the common denominator. 

In this case, the least common denominator, or the least 
number that will contain all the denominators, is 16; hence, 
we must reduce all these fractions to sixteenths and then add 
their numerators. 

13 5 1 

7 - + 3 - + — = ? To reduce the fraction — to a fraction 

4 o lb 4 

having 16 for a denominator, we must multiply both terms 








ARITHMETIC. 


19 


of the fraction by some number which will make the denom¬ 
inator 16. This number evidently is 4, hence, i ^ ^ — 

^ ’ 4 X 4 16' 


Similarly, both terms of the fraction 


— must be multiplied 

o 


3x2 6 

by 2 to make the denominator 16, and we have — =—. 

8 X 2 16 


The fractions now have a common denominator 16; hence, 
we find their sum by adding the numerators and placing their 

4 6 5 

sum over the common denominator, thus: — H- 1 -= 

’ 16 ^ 16 ^ 16 


4 + 6 + 5 _ 15 
16 “ 16* 


Ans. 


(40) When mixed numbers and whole numbers are to be 
added, add the fractional parts of the mixed numbers sep¬ 
arately, and if the resulting fraction is an improper fraction, 
reduce it to a whole or mixed number. Next, add all the 
whole numbers, including the one obtained from the addition 
of the fractional parts, and annex to their sum the fraction 
of the mixed number obtained from reducing the improper 
fraction. 


5 7 5 

42 + 31-r + 9— = ? Reducing -- to a fraction having 

8 16 o 

a denominator of 16, we have X ? Adding the two 

O /i io 

10 7 

fractional parts of the mixed numbers we have J 0 + Jg = 

10 + 7 _ 17 _ 1 
16 16 ^16* 


The problem now becomes 42 + 31 + 9 + 1— = ? 


42 

31 

9 

liV 


16 

Adding all the whole numbers and the 
number obtained from adding the fractional 

parts of the mixed numbers, we obtain 83— 


Ans. as their sum. 





20 


ARITHMETIC. 


(41) 29| + 50| + 41 + G9^ 


3_ 3 X 4^13 
• 4 4X4 16‘ 


6^5X2 
8 8X2 


10 4_ ^ _ 12 -f- 10 -f~ ^ ^ 

Tg* 16'^l6'^l6“ 16 


16 


16 


The problem now becomes 29 -f 50 + 41 -f- 69 + Ij^ — ? 


29 square inches. 

50 square inches. 

41 square inches. 

69 square inches. 

Ij®^ square inches. 

190y^-g- square inches. Ans. 




16 


The line between 7 and — means that 7 is to be divided 
Id 

^^Te- 

^ 3 

32 . 15 . 5_15 3 . 

T “ 32 • 8 “ 32 ^ 5 ” X ^ ~ 4- 
8 4 


4 + 3 7 

(^)^ = l=^ = ro- (See Art. 131.) Ans. 

7 

(43) — = value of the fraction, and 28 = the numerator 

o 

We find that 4 multiplied by 7 = 28, so multiplying 8 , the 
denominator of the fraction, by 4, we have 32 for the required 
28 7 

denominator, and — = —. Hence, 32 is the required de- 

o2 8 

nominator. Ans. 


(44) 


7 7 

8 -l 6 =^ 


When the denominators of frac¬ 


tions are not alike it is evident that the units are divided 
into unequal parts^ therefore, before subtracting, reduce the 









ARITHMETIC. 


21 


fractions to fractions having a common deno 7 ninator. Then, 
subtract the mimerators^ and place the remainder over the 
common denominator. 

7 X 2 _ 14 14 7 _ 14 - 7 

8X2“ 


16 16 16 


16 


=r6- 


(b) 13 - 7-^ = ? This problem may be solved in two 


16 

ways: 

First: 13 = 12^, since ^ = 1, and = 12 + ^ = 
12 + 1 = 13. 

12^|- We can now subtract the whole numbers sepa- 
rately, and the fractions separately, and obtain 12 — 7 


T7 ,16 7 16- 

=5and---= — 


7 9 , 9 _9 

- = I6- ® + i6 = % 


Ans. 


Second: By reducing both numbers to improper fractions 
having a denominator of 16. 


13^13X16^208 7 ^ (7 X 16) + 7_ 112 + 7 

1 1 X16 16* M6 16 ~ 16 


119 

16* 


c, , , , 208 119 208 - 119 89 

Subtracting, we have —- -tt =-- = v: and 

16 16 16 16 


89 

16 


= 16)89(5* 


80 

”9 

16 


the same result that was obtained by the 
first method. 


(r) 312 I - 229| = ? 


We first reduce 


the fractions of the two mixed numbers to 
fractions having a common denominator. Doing this we 
9 9 X 2 18 

have — = —- 77 = 7 : 7 ;. We can now subtract the whole 

16 16 X 2 32 

numbers and fractions separately, and have 312 — 229 = 83 
18-5_13 
32 “ 32* 


'18 5 

32 32 


312^1 

2297®j' 

Isii 












23 


ARITHMETIC. 


The man evidently traveled 85 78 + 125 ^ 

l/v 10 oO 


(45) 

miles. 

Adding the fractions separately in this case, 
5 9 17_ 5,3 17_ 175 + 252 + 204 _ 

12"^15'^35~12'^5"^35“ 420 


631 _ 211 
420“ ^420' 

Adding the whole numbers and the mixed number 85 
representing the sum of the fractions, the sum is 78 

289 miles. Ans. 

To find the least common denominator, we have 
5)12, 5, 35 


289iU 


7)12, 1, 7 


12 , 1 , 


(46) 


573— tons, 
o 

216tons, 
o 


difference 357 ^ tons. Ans. 


1, or 5X7X12 = 420. 

5 “ 40 
5 

8 “40 


— = difference. 


(47) 


Reducing 9 j to an improper fraction, it becomes 


37 AT u- 1 ■ 37 , 3 37 3 111 , 15 , ,, 

—. Multiplying _ by — X g = = 3 — dollars. Ans 

(48) Referring to Arts. 114 and 116, 

^ of ^ of of ^ of 11 multiplied by ^ of ^ of 45 = 

8 6 


11 20 


3 

n 


?x|!x7xl9x;;x7x5x# _ 7x19x7x5x3 _ 13,965 
)!x4x;ix^0xlx8x^xl 4x4x8 ~ 128 “ 

4 ^ 

Ans. 


109 


13 


128‘ 


(49) 

(50) 


3 3 ;0 

7 - of 16 = xX — — 


6 

= 12. 12 -T-1 = l^x|=18. Ans.. 


1 

.,,1,,, 7 845 15 , . , . , 

211 ^ X 1 g = -j- X -g , reducing the mixed numbers 










ARITHMETIC. 


23 


^ . . . 845 15 12,675 

to improper fractions. —j- x - 5 - = —— cents = amount 

4 o oZ 

paid for the lead. The number of pounds sold is evidently 

2,535 

12,675 .1 n.m .A 2,535 7 , _ 

16 


amount remaining is 211 — 


‘-f. 


2,535 845 


16 


13 


16 


52^ pounds. Ans. 
Id 


845 2,535 3,380 


16 


16 


( 51 ) 


^ Xi 

0 8 = Eight hundredths. 


l|i 

S 3 5 

.13 1 = On£ hundred thirty~one thousandths. 


5 

•o 
. a> 
to 

si *0 
0 § 
2 X 
0 0 


5 i 

li 

r 


01 = One ten^thousandth. 


-a 2 -M TJ 


n 3 

0 0 


!3 

2 n g 
5 2 ^ S 

0 0 2 7= Twenty-seven millionths. 


. 0) 


S 


•S a ^ 
o S o fl 
2^22 

0 10 8 = One hundred eight ten-thousandths. 









24 


ARITHMETIC. 


5 


93.0 1 0 1 = Ninety-three, 


and one hundred one ten-tlioasandths 


In reading decimals, read the number just as you would i( 
there were no ciphers before it. Then count from the decimal 
point towards the right, beginning with tenths, to as many 
places as there are figures, and the na7ne of the last figure 
must be annexed to the previous reading of the figures to 
give the decimal reading. Thus, in the first example above, 
the simple reading of the figure is eighty and the name of its 
position in the decimal scale is liundredths, so that the 
decimal reading is eight liundredths. Similarly, the fig¬ 
ures in the fourth example are ordinarily read twenty-seven ; 
the name of the position of the figure 7 in the decimal scale 
is millionths, giving, therefore, the decimal reading as 
twenty-seven millionths. 

If there should be a whole number before the decimal 
point, read it as you would read any whole number, and 
read the decimal as you would if the whole number were 
not there; or, read the whole number and then say, and ” 
so many hundredths, thousandths, or whatever it may be, 
as “ ninety-three, and one hundred one ten thousandths.” 


(52) See Art. 139. 

(53) See Art. 153. 

(54) See Art. 160. 

(55) A fraction is one or more of the equal parts of a 
unit, and is expressed by a numerator and a denominator, 
while a decimal fraction is a number of tenths^ hundredths^ 
thousandths^ etc., of a unit, and is expressed by placing a 
period (.), called a decimal point, to the left of the figures 
of the number, and omitting the denominator. 


(56) See Art. 165. 


ARITHMETIC. 


25 


( 57 ) 


To reduce the fraction ^ to a decimal, we annex 
2 


one cipher to the numerator, which makes it 1.0. Dividing 
1.0, the numerator, by 2, the denominator, gives a quotient 
of .5, the decimal point being placed before the figure 
of the quotient, or .5, since only one cipher was annexed to 
the numerator. Ans. 


8)7.000 

.875 


Ans. 


32)5.00000(.15625 Ans. 
32 


Since . 65 = 


65 


65 

100 - 

must equal .65. Or, when 
the denominator is 10, 100, 
1000, etc., point off as many 
places in the numerator as 
there are ciphers in the 
denominator. Doing so, 
65 


100 


= .65. Ans. 


180 

160 


200 
192 • 


125 

•1000 


= .125. Ans. 


80 

64 

160 

160 


( 58 ) {a) This example, written in the form of a fraction, 

means that the numerator (32.5 + -^9 + 1-5) is to be divided 
by the denominator (4.7 + 9). The operation is as follows: 
32.5+ .29 + 1.5 
4 7 + 9 

3 2.5 
+ .29 

+ 1.5 


13.7 ) 3 4.2 9 000 ( 2.5 02 9 Ans. 



274 

Since there are 5 deci¬ 

4.7 

689 

mal places in the dividend 

+ 9.0 

685 

and 1 in the divisor, there 


400 

are 5 — 1 or 4 places to 


274 

be pointed off in the quo¬ 


1260 

tient. The fifth figure of 


1233 

the decimal is evidently 


27 

less than 5. 


11. a. IV.— 3 












26 


ARITHMETIC. 


(d) Here again the problem is to divide the numerator, 
which is (1.283 X 8 + 5), by the denominator, which is 2.63. 
The operation is as follows: 


1.283 X 8 + 5 
2.63 


= ? 8 + 5: 

1.2 8 3 
X 13 
3849 
1283 


:13. 


2.63)16.6 7 9 000(6.3 418 Ans. 
1578 





899 

789 


480 

263 


1100 


2170 


1052 


2104 


480 

163 
- 8 

155 

66 

589 + 27 X 163 
25 + 39 



589 
+ 27 


X 616 

930 

155 


He 


930 



64)95480.0 00(1491.873 


25 
+ 3 9 

64 


There are three deci¬ 
mal places in the quotient, 
since three ciphers were 
annexed to the dividend. 


64 
314 
256 
588 
576 


Ans. 


120 

64 

560 

512 


480 

448 


320 

320 






















ARITHMETIC. 


27 


(d) 


40.6 +7.1 X (3.039 - 1.874) 

6.27 + 8.53- 8.01 “ 

40.6 

3.03 9 

+ ^ 

- 1.87 4 

47.7 

1.15 5 


X 47.7 

6.2 7 

8 08 5 

4- 8.5 3 

8 0 85 

14.80 

4 6 2 0 

- 8.01 

6.79)5 5.0 935 

6.79 

5 4 3 3 


6 decimal places in 
the dividend — 2 deci¬ 
mal places in the divi¬ 
sor = 4 decimal places 
to be pointed off in 
the quotient. 


7 73 
6 79 
945 
679 

2660 

2037 

6230 

6111 


(59) .875 = 

1 foot = 12 inches. 


119 

875 175 7 . , , 

1,000 200 8 ® 


7 7 12 21 1 

g of 1 foot = -g X Y = "2 = lO^ inches. Ans. 


(60) 12 inches = 1 foot. 

lofan inch = 1-12 = 1 = ^ of a foot. 

4 

Point off 6 decimal places in the quotient, since we 
annexed six ciphers to the dividend, the divisor con¬ 
taining no decimal places; hence, 6 — 0=6 places to be 
pointed off. 















28 


ARITHMETIC. 


1 


64 ) 1.000000(.015626 Ans. 

64 

360 

320 

~400 

384 

160 

128 

320 

320 

( 61 ) If 1 cubic inch of water weighs .03617 of a pound, 
the weight of 1,500 cubic inches will be .03617 X 1,500 — 
54.255 lb. 

.03617 lb. 

1500 

1808500 

3617 

5 4.25500 lb. Ans. 

( 62 ) 72.6 feet of fencing at $.50 a foot would cost 

7 2.6 X .50, or $36.30. 

.50 

$3 6.3 0 0 

If. by selling a carload of coal at a profit of $1.65 per ton, 
I make $36.30, then there must be as many tons of coal in 
the car as 1.65 is contained times in 36.30, or 22 tons. 

1.6 5)3 6.30( 2 2 tons. Ans. 

330 


330 

330 









ARITHMETIC. 


29 


(63) 231 ) 17892.00000 ( 77.45454, or77.4545to 

1617 four decimal places. Ans 

17 22 
1617 

1050 

924 

1260 

1155 

1050 

924 

1260 

1155 

1050 

37.13 % .0952 

X X X 19 X 19 X 350 _ 

X X ^ 

1,000 % 


37.13 X 

.0952 xl9xl9 x 350 

446,618.947600 

1,000 1,000 

446.619 to three decimal places. Ans. 

37.13 

19 

361 

3.534776 

.0952 

19 

350 

126350 

7426 

171 

18050 

176738800 

18565 

19 

1083 

10604328 

33417 

3^ 

126350 

21208656 

3.534776 

7069552 

3534776 


446618.947600 


(65) See Art. 174. Applying rule in Art. 175<, 


(a) .7928 = 

QO 

{b) .1416 X^ = 

(c) .47915 X^ = 


64 50.7392 


4.5312 

32 

7.6664 

16 


= & 

= |. Ans. 

8 1a 

:-=^. Ans. 



















30 


ARITHMETIC. 


( 66 ) In subtraction of decimals, 
place the decimal points directly 
under each other^ and proceed as in 
the subtraction of whole numbers, 


709.6300 

.8514 

708.7786 Ans. 


placing the decimal point in the remainder directly under 
the decimal points above. 

In the above example we proceed as follows: We can not 
subtract 4 ten-thousandths from 0 ten-thousandths, and, as 
there are no thousandths, we take 1 hundredth from the three 
hundredths. 1 hundredth = 10 thousandths — 100 ten-thou¬ 
sandths. 4 ten-thousandths from 100 ten-thousandths leaves 
96 ten-thousandths. 96 ten-thousandths = 9 thousandths -J- 6 
ten-thousandths. Write the 6 ten-thousandths in the ten- 
thousandths place in the remainder. The next figure in the 
subtrahend is 1 thousandth. This must be subtracted from 
the 9 thousandths which is a part of the 1 hundredth taken 
previously from the 3 hundredths. Subtracting, we have 1 
thousandth from 9 thousandths leaves 8 thousandths, the 8 
being written in its place in the remainder. Next we have 
to subtract 5 hundredths from 2 hundredths (1 hundredth 
having been taken from the 3 hundredths makes it but 2 
hundredths now). Since we can not do this, we take 1 tenth 
from 6 tenths. 1 tenth ( = 10 hundredths) + ^ hundredths 
= 12 hundredths. 5 hundredths from 12 hundredths leaves 
7 hundredths. Write the 7 in the hundredths place in the 
remainder. Next we have to subtract 8 tenths from 5 tenths 
(5 tenths now, because 1 tenth was taken from the 6 tenths). 
Since this can not be done, we take 1 unit from the 9 units. 
1 unit — 10 tenths; 10 tenths 5 tenths = 15 tenths, and 8 
tenths from 15 tenths leaves 7 tenths. Write the 7 in the 
tenths place in the remainder. In the minuend we now have 
708 units (one unit having been taken away) and 0 units in the 
subtrahend. 0 units from 708 units leaves 708 units; hence, 
we write 708 in the remainder. 


{J}) 81.9 6 3 

1.700 


(^) 


80.2 6 3 Ans. 


18.0 0 
.18 

1 7.82 Ans. 


(^) 


1.000 

.001 

.9 9 9 Ans. 






ARITHMETIC. 


31 


(e) 872.1 - (. 8721 + . 008) = ? In this prob. 

lem we are to subtract (.872.1 + .008) from 
872.1. First perform the operation as indi- 
cated by the sign between the decimals .SSOl S2i7n. 
enclosed by the parenthesis. 

Subtracting the sum (obtained by adding the decimals 


872.1000 

.8801 

8 7 1.2199 Ans. 


enclosed within the parenthesis) from 
the number 872.1 (as required by the 
minus sign before the parenthesis), 
we obtain the required remainder. 

First perform 


(/) (5.028 + .0073) - (6.704 - 2.38) =? 

the operations as indicated by the signs be¬ 
tween the numbers enclosed by the paren¬ 
theses. The first parenthesis shows that 
5.028 and .0073 are to be added. This 
gives 5.0353 as their sum. 

6.704 
2.380 


5.02 80 
.0073 

5.0 3 5 3 sum. 


4.3 2 4 difference. 
The sign between 

5.03 5 3 
4.3 2 4 


The second parenthesis shows that 
2.38 is to be subtracted from 6.704. 
The difference is found to be 4.324. 
the parentheses indicates that the 
quantities obtained by performing 
the above operations, are to be sub¬ 
tracted, namely, that 4.324 is to be 
subtracted from 5.0353. Perform- 


.7113 Ans. 

ing this operation we obtain .7113 as the final result. 


(67) In subtracting a decimal from a fraction, or sub¬ 
tracting a fraction from a decimal, either reduce the fraction 
to a decimal before subtracting, or reduce the decimal to a 
fraction and then subtract. 


W |.-.807 = ? 


.875 

.807 

.0 6 8 Ans. 


reduced to a decimal becomes 


7 

8 )7.000 

.875 


Subtracting .807 from .875 the re¬ 
mainder is .068, as shown. 









32 


ARITHMETIC. 


(^) 
.875 = 


.875 — —=? Reducing .875 to a fraction we have 
8 


875 175 35 7 ^ 7 

1,000 "" 200 "40“ 8’ 8 


3_7-3_4_l 
8 " 8 " 8 " 2 * 

3 3 Ans. 

Or, by reducing g- to a decimal, — ^ g qqq and then sub¬ 


tracting, we obtain .875 

i, the same answer as above. 
2 


.375 

.375 = .5 = 4= 

10 .3 7 5 


.5 0 0 Ans. 


(c) ^4 + .435^ — — .07^ = ? We first perform the 

operations as indicated by the signs-between the numbers 

• 5 

enclosed by the parentheses. Reduce — to a decimal and 

5 

we obtain = .15625 (see example 57). 

O/v 


Adding .15625 and .435, 


.15625 

.435 


sum .59125 


21 

j—= .21; subtracting, .21 
.07 

difference .14 


We are now prepared to perform the .5 9125 
operation indicated by the minus sign be- *14 
tween the parentheses, which is, difference .45125 Ans. 

(<r/) This problem means that 33 millionths and 17 thou¬ 
sandths are to be added. Also, that 53 hundredths and 
274 thousandths are to be added, and the smaller of these 
sums is to be subtracted from the larger sum. Thus, 
(. 53 + .274) - (. 000033 4 - .017) = ? , 


t/5 

4> 



x: 

-d 

c 



.0 0 0 0 3 3 
.0 1 7 

.0 1 7 0 3 3 sum. 


.804 larger sum, 
."Ss .017033 S7naller sunu 

w u ^ _ 

| 1 | difference .786967 Ans. 

4-» fC +J 

.5 3 
.2 7 4 
.8 0 4 su7n. 











ARITHMETIC. 


33 


( 68 ) In addition of decimals the 
decimal points must he placed directly 
under each other, so that tenths will 
come under tenths, hundredths under 
hundredths, thousandths under thou¬ 
sandths, etc. The addition is then 
performed as in whole numbers, the 
decimal point of the sum being placed 
directly under the decimal points above. 


.125 

.7 

.089 

.4005 

.9 

.000027 


Ans. 


(69) 9 2 7.416 

8.2 7 4 
3 7 2.6 
62.0 7 9 38 

1370.36938 Ans. 


(70) 


(U 


xh 

a 

CO q 

*§ o 

g-s 


Cl S o d g 
<0 2 <0 2 


.017 


.000047 


.217047 = Ans. 


(71) (a) There 
.107 
.013 

321 

107 

.001391 Ans. 

(b) 2 03 
2.0 3 

eTo 

406 0 

412.09 
.2 03 

123627 
824180 
83.65427 Ans^ 


are 3 decimal places in the multi“ 
plicand and 3 in the multiplier; hence, 
there are 3 + 3 or 6 decimal places in 
the product. Since the product con¬ 
tains but four figures, we prefix two 
ciphers in order to obtain the neces¬ 
sary six decimal places. 

There are two decimal places in the 
multiplier and none in the multipli¬ 
cand ; hence, there are 2 + 0 or two 
decimal places in the first product. 

Since there are 2 decimal places in 
the multiplicand and 3 decimal places 
in the multiplier, there are 3 2 or 5 

decimal places in the second product. 


) 










34 


ARITHMETIC. 


(c) First perform the operations indicated by the signs 
between the numbers enclosed by the parenthesis, and then 
perform whatever may be required by the sign before the 
parenthesis. 

Multiply together the numbers 2.7 
and 31.85. 

The parenthesis shows that .316 is 
to be taken from 3.16. 3.16 0 

.316 


2.844 


31.85 

2.7 

22295 

6370 

85.995 


The product obtained by the first 
operation is now multiplied by the 
remainder obtained by performing 
the operation indicated by the signs 
within the parenthesis. 


85.995 

2.844 

343980 

343980 

687960 

171990 


244.569780 Ans. 


(d) (107.8 + 6.541 - 31.96) X 1.742 = ? 

107.8 
+ 6.541 

114.341 
— 31.96 

8 2.3 81 
X 1.742 

164762 

329524 

576667 

82381 

14 3.507702 Ans. 


(72) {a) — .13^ X .625 + — — ? 

First perform the operation indicated by the parenthesis. 












ARITHMETIC. 


35 


1 = 1 

16 16)7.0000(.4375 

64 

48 

120 

112 

80 


We point off four decimal 
places since we annexed four 
ciphers. 


.4375 

.13 

Subtracting, we obtain .3075 

The vinculum has the same meaning as the parenthesis; 
5 _ 5 hence, we perform the operation indicated 

8 “ 8 ) Q Q by it. We point off three decimal places, 
.6 25 since three ciphers were annexed to the 5. 

Adding the terms in- .6 2 5 
eluded by the vinculum, .6 2 5 
we obtain 1.2 5 0 

The final operation is to perform the work indicated by 
the sign between the parenthesis and the vinculum, thus, 

.3075 

1.25 


(»)(gx.Sl)-(^02xi) = 


15375' 

6150 

3075 

.384375 Ans. 
? 


01-^1 19 21 ^ 399 

'^^“100’ 32^100 


. 02 = 


^ _3_ X 4 
800 800 X 4 


3200 

12 399 

3200 


2 

100 ' 


Axl 

100 16 


6 


3200 


12 

3200 


^ 3 

1600 800' 

399 - 12 _ 387 
3200 “ 3200' 


3200 













36 


ARITHMETIC. 


007 

Reducing J200 


to a decimal, we obtain 


3200 ) 38 7.0 000000 (.1 209375 Ans. 
3 2 0 0 

6 700 
6400 


30000 

28800 

12 0 0 0 
9600 

24000 

22400 

16000 

16000 


Point off seven decimal 
places, since seven ciphers 
were annexed to the divi¬ 
dend. 


(c) (^ + .013-2.17)xl3l-7A=? 

13 _ 13 Point off two decimal 

4 4)13.00 places, since two ciphers 

3.2 5 were annexed to the divi¬ 
dend. 

g 

— reduced to a decimal is .3125, since 


3.2 5 
+ .013 

3.2 6 3 
-2.17 

1.093 


16)5.0000(.3125 Point off four decimal 
^ ^ places, since four ciphers 

2 0 were annexed to the 

16 dividend. 

40 

32 

80 

80 


Then, 7^ = 7.3125, and 13j = 13.25, since 


1 _ 1 

4 “ 4 ) 1,00 


.25 














ARITHMETIC. 


37 


13.2 5 5.9 3 75 

- 7-3125 X 1.0 9 3 

5.9375 178125 

534375 

593750 

6.489 6 87 5 Ans. 

(73) (^) .875 i =.875 -7-.5 ^since =.5^ = 1.75. Ans. 

Another way of solving this is to reduce .875 to its equivalent 
common fraction and then divide. 


875 175 35 7 

.875 = —, since .875 =-- =-= — = - • 

8’ 1,000 200 40 8’ 


1 = lv?. = L=ii 

2 ^ ^ 1 4 ^ 4 * 

4 


I = 1)3.00(.75, 


the same answer as above. 


28 

20 

20 


then, g--^- 
l| = 1.75, 


(b) -^.5 = -^-(smce.5 = g-j=^xf = j = lj, 
1.75. Ans. . 4 


or 


This can also be solved by reducing — to its equivalent 

8 

decimal and dividing by .5;'| = .875; .875 -^.5 = 1.75. 

Since there are three decimal places in the dividend and one 
in the divisor, there are 3 -- 1, or 2 decimal places in the 
quotient. 

We shall solve this problem by first 
reducing the decimals to their equiva¬ 
lent common fractions. 

_ 15_3 
“ “ 8 * 


,375_2i.i^, 

^ ^ A-.125 


.375 = 


375 


75 


1,000 200 40 

the numerator of the fraction. 
125 _ 25 _ 1 
~ 200 “ 


3 13 

g X ^ or the value of 


125 = 


1,000 


8 


Reducing — to sixteenths, we 
8 











38 


ARITHMETIC. 


\ 


nominator of the fraction. The problem is now reduced to 
3 3 


32 

3 


? 


16 



16 “ ^ ^ “ 2 
2 


or .5. 


Ans. 


. . 1.25 X 20 X 3 ^ In this problem 1.25 X 20 X 

' ' 87 + (11 X 8) * 3 constitutes the numerator of 

459 + 32 the complex fraction. 


1.2 5 Multiplying the factors of the numerator 
X 2 0 together, we find their product to be 75. 

25.00 
X ^ 

*75 


87 (11 X 8) 

The fraction •— i ~ qq ~~ " ' constitutes the denominator of 
4:Oy O/i 

the complex fraction. The value of the numerator of this 
fraction equals 87 + 88 = 175. 

The numerator is combined as though it were written 
87 + (11 X 8), and its result is 


11 

8 times 


88 
+ 87 

175 

The value of the denominator of this fraction is equal to 
459 + 32 = 491. The problem then becomes 

3 

75 75.175 75 ^491 7^x 491 1,473 .,.3 ^ 

491 7 


(75) 1 plus .001 = 1.001. .01 plus .000001 = .010001. 

And 1.001 - .010001 = 

1.001 

.010001 


.990999 Ans. 









ARITHMETIC 

(SECTION 4.) 

(QUESTIONS 76-117.) 

( 76 ) A certain per cent, of a number means so many 
hundredths of that number. 

25^ of 8,428 lb. means 25 hundredths of 8,428 lb. Hence, 
25^ of 8,428 lb. = .25 X 8,428 lb. = 2,107 lb. Ans. 

( 77 ) Here $100 is the base and 1^ = .01 is the rate. 
Then, .01 X $100 = $1. Ans. 

(78) means one-half of one per cent. Since is 
Z 

.01, is .005, for, ^ . And .005 X 135,000 = $175. 

^ -005 Ans. 

( 79 ) Here 50 is the base, 2 is the percentage, and it is 
required to find the rate. Applying rule. Art. 193 , 

rate = percentage ~ base; 
rate = 2 -E 50 = .04 or 4^. Ans. 

( 80 ) By Art. 193 , rate = percentage base.* 

As percentage = 10 and base = 10, we have rate = 10 
10 = 1 = 100^. Hence, 10 is 100^ of 10. Ans. 

( 81 ) (a) Rate = percentage-f-by base. Art. 193 . 

As percentage = $176.54 and base = $2,522, we have 

rate = 176.54 -e 2,522 = .07 = 7^. Ans. 

2 5 2 2 ) 1 7 6.54 
.07 


* Remember that an expression of this form means that the first 
term is to be divided by the second term. Thus, as above, it means 
percentage divided by base. 

For notice of the copyright, see page immediately following the title page. 






40 


ARITHMETIC. 


(d) Base = percentage rate. Art. 192. 

As percentage = 16.96 and rate = 8^ = .08, we have 

base = 16.96 ~ .08 = 212, Ans. 

.08 ) 1 6.9 6 
2 12 

(^) Amount is the sum of the base and percentage; hence 
the percentage = amount minus the base. 

Amount = 216.7025 and base = 213.5; hence, percentage = 
216.7025 - 213.5 = 3.2025. 

Rate = percentage ~ base. Art. 193. 

Therefore, rate = 3.2025 ~ 213.5 = .015 = 1^^. Ans. 

213.5 ) 3.202 5 ( .015 = li^ 

2135 

10 6 7 5 
10675 


(d) The difference is the remainder found by subtracting 
the percentage from the base; hence, ba^e — the differ¬ 
ence = the percentage. Base = 207 and difference =201.825, 
hence percentage = 207 — 201.825 = 5.175. 

Rate = percentage ~~ base. Art. 193. 

Therefore, rate = 5.175 307 = .035 = .02^ = 2^^. Ans. 

2 2 ■ 

207 ) 5.175 ( .025 
4 14 

103 5 
1035 


(82) In this problem $5,500 is the amount, since it 
equals what he paid for the farm + what he gained; 
15^ is the rate, and the cost (to be found) is the base. 
Applying rule. Art. 197, 

base = amount (1 + rate); hence, 

base = $5,500 -f- (1 + .15) = $4,782.61. Ans. 







ARITHMETIC. 


41 


1.15)5500.0000 ( 4782.61 
460 

9 00 
805 

950 

920 

300 

230 

700 

690 

100 

115 

The example can also be solved as follows: 100^ = cost; if he 
gained 15^, then 100+ 15= 115^ = 15,500, the selling price. 

If 115;^ = $5,500, = + of $5,500 = $47.8261, and 100.!^, 

lio 

or the cost, = 100 X $47.8261 = $4,782.61. Ans. 

( 83 ) 24 ^ of $950 = .24 X 950 = $228 

12-|^of $950 = .125 X 950= 118.75 

17 i of $950 = .17 X 950 = 161.50 

53-1 of $950 = $508.25 

The total amount of his yearly expenses, then, is $508.25, 
hence his savings are $950 — $508.25 = $441.75. Ans. 

Or, as above, 24^ + 12 ^^ + 17^ = 53^^, the total per- 
2 2 

centage of expenditures; hence, 100^ — 53^^ = 46^^ = pet 

2 2 

cent, saved. And $950 X .465 = $441.75 = his yearly sav¬ 
ings. Ans. 

(84) The percentage is 961.38, and the rate is.37 By 

Art. 192, 

Base = percentage -f- rate 

= 961.38 .375 = 2,563.68, the number. Ans. 


B.O. IV.—4 









42 


ARITHMETIC. 


Another method of solv¬ 
ing is the following: 

If 37 of a number is 

961.38, then .37^ times the 

number = 961.38 and the 

number = 961.38 -37^, 

which, as above = 2,563.68. 

Ans. 


.375)961.38000(2563.68 

750 

2113 

1875 

2388 

2250 

1380 

1125 

2 5 50 
2250 

3000 

3000 


( 85 ) Here $1,125 is 30^ of some number; hence, 
$1,125 = the percentage, 30^ = the rate, and the required 
number is the base. Applying rule. Art. 192 , 

Base = percentage -v- rate = $1,125 ~ .30 = $3,750. 


3 1 

Since $3,750 is j of the property, one of the fourths is ~ 


of $3,750 = $1,250, and — or the entire property, is 4 X $1,250 
= $5,000. Ans. 


(86) Here $4,810 is the difference and 35^ the rate. By 
Art. 198 , 

Base = difference -j- (1 — rate) 

= $4,810 ^ (1 - .35) = $4,810 .65 = $7,400. Ans. 

.65)4810.00(7400 

455 

260 1.00 

260 .35 

00 ■ .65 

Solution can also be effected as follows: 100^ = the sum 
diminished by 35^, then (1 — .35) = .65, which is $4,810. 











ARITHMETIC. 


43 


If 65^ = $4,810, l^ = ^oi 4,810 = $74, and 100^ = 100 X 

DO 

$74 = $7,400. Ans. 

( 87 ) In this example the sales on Monday amounted to 

$197.55, which was 12^^ of the sales for the entire week; 

i. e., we have given the percentage, $197.55, and the rate, 

12^^, and the required number (or the amount of sales for 
Z 

the week) equals the base. By Art. 192, 

Base = percentage rate = $197.55 -f- .125; 
or, .125)197.5500(1580.4 Ans. 

125 

725 

625 

1005 

1000 

500 

Therefore, base = $1,580.40, which also equals the sales 
for the week. 

(88) 16.5 miles = 12^^ of the entire length of the road. 
We wish to find the e/itire length. 

16.5 miles is the percentage, 12^^ is the rate, and the en- 

tire length will be the base. By Art. 192, 

Base = percentage rate = 16.5 

.125) 16.500(132 miles. Ans. 

12 5 

4 00 
3 75 


250 
2 59 







44 


ARITHMETIC. 


( 89 ) Here we have given the difference, or $35, and the 
rate, or 60^, to find the base. We use the rule in Art. 198 , 
Base = difference (1 — rate) 

= $35 (1 — .60) = $35 .40 = $87.50. Ans. 

.40)35.000(87.5 
32 0 

3 00 
2 80 

200 

200 


Or, 100^ = whole debt; 100^ — 60^ = 40^ = $35. 

1 35 

If 40^ = $35, then H = of $35 = and 100^ = 

OK 

100 = $87.50. Ans. 


( 90 ) 28 rd. 4 yd. 2 ft. 10 in. to inches. 


X_ 5i 

154 

+_f 

15 8 yards 
X 3 

4 ^ 

+ 2 

4 7 6 feet 
X 12 

5712 
+ 10 

5 7 22 inches. Ans. 


Since there are 5^ yards in 
one rod, in 28 rods there are 
28 X 5^- or 154 yards; 154 yards 
plus 4 yards = 158 yards. There 
are 3 feet in one yard; there¬ 
fore, in 158 yards there are 
3 X 158 or 474 feet; 474 feet + 
2 feet = 476 feet. There are 
12 inches in one foot, and in 
476 feet there are 12 X 476 or 
5,712 inches; 5,712 inches -f 10 
inches = 5,722 inches. Ans. 


( 91 ) 12 ) 5 7 2 2 inches. 

3 ) 4 7 6 -|- 10 inches. 
5^ ) 158 + 2 feet. 

2 8 + 4 yards. 

Ans. = 28 rd. 4 yd. 2 ft. 10 in. 









arithmetic. 


45 


Explanation. —There are 12 inches in 1 foot; hence, in 
5,722 inches there are as many feet as 12 is contained times 
in 5,722 inches, or 476 ft. and 10 inches remaining. Write 
these 10 inches as a remainder. There are 3 feet in 1 yard; 
hence, in 476 feet there are as many yards as 3 is contained 
times in 476 feet, or 158 yards and 2 feet remaining. There 

are 5-^ yards in one rod; hence, in 158 yards there are 28 rods 

and 4 yards remaining. Then, in 5,722 inches there are 
28 rd. 4 yd. 2 ft. 10 in. 

( 92 ) 5 weeks 3.5 days. 

X_7 

3 5 days in 5 weeks. 

+ 3.5 

3 8.5 days. 

Then, we find how many seconds there are in 38.5 days. 

3 8.5 days 

X 2 4 hours in one day. 

1540 

770 

9 2 4.0 hours in 38.5 days. 

X 6 0 minutes in one hour. 

5 5 440 minutes in 38.5 days. 

X 6 0 seconds in one minute. 

3 3 2 6 4 0 0 seconds in 38.5 days. Ans. 

( 93 ) Since there are 24 gr. in 1 pwt., in 13,750 gr. there 
are as many pennyweights as 24 is contained times in 
13,750, or 572 pwt. and 22 gr. remaining. Since there are 
20 pwt. in 1 oz., in 572 pwt. there are as many ounces as 
20 is contained times in 572, or 28 oz. and 12 pwt. remaining. 

Since there are 12 oz. in 1 lb. (Troy), in 28 oz. there are 
as many pounds as 12 is contained times in 28, or 2 lb. and 
4 oz. remaining. We now have the pounds and ounces 
required by the problem; therefore, in 13,750 gr. there are 
2 lb. 4 oz. 12 pwt. 22 gr. 






4G 


ARITHMETIC. 


24 ) 1 3750 gr. 

20 ) 5 7 2 pwt. + 22 gr. 

12 oz. + 12 pwt. 

2 lb. + 4 oz. 

Ans. = 2 lb. 4 oz. 12 pwt. 22 gr. 

( 94 ) 100 ) 47G3254 li. 

80 ) 4 7 G 3 2 + 54 li. 

5 9 5 + 32 ch. 

Ans. = 595 mi. 32 ch. 54 li. 

Explanation. —There are 100 li. in one chain; hence, in 
4,763,254 li. there are as many chains as 100 is contained 
times in 4,763,254 li., or 47,632 ch. and 54 li. remaining. 
Write the 54 li. as a remainder. There are 80 ch. in one 
mile; hence, in 47,632 ch. there are as many miles as 80 is con¬ 
tained times in 47,632 ch., or 595 miles and 32 ch. remaining. 

Then, in 4,763,254 li. there are 595 mi. 32 ch. 54 li. 

( 95 ) 1728 ) 7 64325 cu. in. 

27 ) 442 + 549 cu. in. 

1 6 cu. yd. + 10 cu. ft. 

Ans. = 16 cu. yd. 10 cu. ft. 549 cu. in. 

Explanation. —There are 1,728 cu. in. in one cubic foot; 
hence, in 764,325 cu. in. there are as many cubic feet as 
1,728 is contained times in 764,325, or 442 cu. ft. and 
549 cu. in. remaining. Write the 549 cu. in. as a remainder. 
There are 27 cu. ft. in one cubic yard; hence, in 442 cu. ft. 
there are as many cubic yards as 27 is contained times in 
442 cu. ft., or 16 cu. yd. and 10 cu. ft. remaining. Then, in 
764,325 cu. in. there are 16 cu. yd. 10 cu. ft. 549 cu. in. 

( 96 ) We must arrange the different terms in columns, 
taking care to have like denominations in the same column. 

rd. yd. ft. in, 

2 2 2 3 

4 19 

2 7 


3 2i 0 7 

or 3 2 2 1 Ans. 









ARITHMETIC. 


47 


Explanation.— We begin to add at the right-hand col¬ 
umn. 7 9 3 = 19 in. *, as 12 in. make one foot, 19 in. = 

1 ft. and 7 in. Place the 7 in. in the inches column, and 
reserve the 1 ft. to add to the next column. 

1 (reserved) -}- 2 -[- 1 + = 0 ft. Since 3 ft. make 1 yard, 

6 ft. = 2 yd. and 0 ft. remaining. Place the cipher in the 
column of feet and reserve the 2 yd. for the next column. 

2 (reserved) -f 4 + 2 = 8 yd. Since 5-^ yd. = 1 rod, 8 yd. = 

1 rd. and 2— yd. Place 2^ yd. in the yards column, and 
2 2 


reserve 1 rd. for the next column; 1 (reserved) -j- 2 = 3 rd. 


Ans. 

= 3 rd. 

2iyd. 

0 ft. 

7 in. 

or, 

3 rd. 

2 yd. 

1 ft. 

13 in. 

or, 

3 rd. 

2 yd. 

2 ft. 

1 in. 


gal. 

3 
6 

4 


qt. 

3 

0 

0 

8 


16 gal. 3 qt. 0 pt. 2 gi. 


Ans. 

( 97 ) We write the compound numbers so that the units 
of the same denomination shall stand in the same column. 
Beginning to add with the lowest denomination, we find that 

the sum of the gills is 1 2 + 

3 = 6. Since there are 4 gi. in 

1 pint, in 6 gi. there are as many 
pints as 4 is contained times in 
6, or 1 pt. and 2 gi. We place 

2 gi. under the gills column 
and reserve the 1 pt. for the 
pints column; the sum of the 

pints is 1 (reserved) -}-5 + l + l = 8. Since there are 2 pt. 
in 1 quart, in 8 pt. there are as many quarts as 2 is con¬ 
tained times in 8, or 4 qt. and 0 pt. We place the cipher 
under the column of pints and reserve the 4 for the quarts 
column. The sum of the quarts is 4 (reserved) 8 + 3 = 15. 
Since there are 4 qt. in 1 gallon, in 15 qt. there are as many 
gallons as 4 is contained times in 15, or 3 gal. and 3 qt. re¬ 
maining. We now place the 3 under the quarts column 
and reserve the 3 gal. for the gallons column. The sum of 
the gallons column is 3 (reserved) + 4+ 6 + 3 = 16 gal. 
Since we can not reduce 16 gal. to any higher denomination, 
we have 16 gal. 3 qt. 0 pt. and 2 gi. for the answer. 



48 


ARITHMETIC. 


(98) Reduce the grains, pennyweights, and ounces to 
higher denominations. 

24) 240 gr. 20 ) 125 pwt. 12 ) 50 oz. 

10 pwt. 6 oz. 5 pwt. 4 lb. 2 oz. 

Then, 3 lb. + 4 lb. 2 oz. -j- 6 oz. 5 pwt. -f 10 pwt. = 
lb. oz. pwt. 

3 

4 2 

6 5 

. 10 

7 lb. 8 oz. 15 pwt. Ans. 

( 99 ) Since “seconds” is the lowest denomination in 

this problem, we find their sum first, which is 11 + 29 + 25 + 

30 -j- 12, or 107 seconds. Since 
there are 60 seconds in 1 minute, 
in 107" there are as many minutes 
as 60 is contained times in 107, or 
1 minute and 47 seconds remain¬ 
ing. We place the 47 under the 
seconds column and reserve the 1 
for the minutes column. The sum 
of the minutes is 1 (reserved) -f 
17 + 26 -{- 19 + 16, or 79. Since there are 60 minutes in 
1 degree, in 79 minutes there are as many degrees as 60 is 
contained times in 79, or 1 degree and 19 minutes remaining. 
We place the 19 under the minutes column and reserve the 

1 degree for the degrees column. The sum of the degrees 
is 1 (reserved) + 10 + 20 + 13 -f 11, or 55 degrees. Since 
we can not reduce 55 degrees to any higher denominations, 
we have 55° 19' 47" for the answer. 

( 100 ) Since “inches” is the lowest denomination in 
this problem, we find their sum first, which is 11 + 8 + 6, or 
25 inches. Since there are 12 inches in 1 foot, in 25 inches 
there are as many feet as 12 is contained times in 25, or 

2 feet and 1 inch remaining. Place the 1 inch under the 
inches column, and reserve the 2 feet to add to the column 


deg. 

min. 

sec. 

11 

16 

12 

13 

19 

30 

20 

0 

25 

0 

26 

29 

10 

17 

11 

55° 

19' 

47" 






ARITHMETIC. 


49 


rd. 

130 

215 

304 


yd. 

5 

0 

4 


in. 

6 

8 

11 


of feet. The sum of the feet is 2 feet (reserved) -|- 2 + 1 = 

5 feet. Since there are 3 
feet in 1 yard, in 5 feet 
there are as many yards as 3 
is contained times in 5 feet, 
or 1 yard and 2 feet remain- 
650 4i 2 1 ing. Place the 2 feet under 

mi. the column of feet, and re- 

or, 2 10 5 0 7 Ans. serve the 1 yard to add to the 

column of yards. The sum of 
the yards is 1 yard (reserved) 4 -f 5 = 10 yards. Since there 

are 5^ yards in 1 rod, in 10 yards there are as many rods as 

5^ is contained times in 10, or 1 rod and 4^ yards remaining. 
2 2 

Place the 4^ yards under the column of yards, and reserve 
2 


the 1 rod for the column of rods. The sum of the rods is 1 
(reserved) + 304 215 + 130 = 650 rods. Place 650 rods 

under the column of rods. Therefore, the sum is 650 rd. 

4^ yd. 2 ft. 1 in. Or, since ^ a yard = 1 ft. 6 in., and since 
2 2 

there are 320 rods in 1 mile, the sum may be expressed as 
2 mi. 10 rd. 5 yd. 0 ft. 7 in. Ans. 

(lOl) vSince “square links ” is the lowest denomination 
in this problem, we find their sum first, which is 21 + 23 

+ 16 + 18 + 23 + 21, or 
122 square links. Place 122 
square links under the col¬ 
umn of square links. The 
sum of the square rods is 

2 -(” 3 “h ^ “I" ^ ^ ~h 

14 square rods. Place 14 
square rods under the col¬ 
umn of square rods. The 
sum of the square chains 
is 323 square chains. Since there are 10 square chains in 
1 acre, in 323 square chains there are as many acres as 10 is 


A. 

sq. ch. 

sq. rd. 

sq. li. 

21 

67 

3 

21 

28 

78 

2 

23 

47 

6 

2 

18 

56 

59 

2 

16 

25 

38 

3 

23 

46 

75 

2 

21 

255 

3 

14 

122 




50 


ARITHMETIC. 


contained times in 323 square chains, or 32 acres and 3 square 
chains remaining. Place 3 square chains under the column 
of square chains, and reserve the 32 acres to add to the col¬ 
umn of acres. The sum of the acres is 32 acres (reserved) -{- 
46 + 25 + 56 -I- 47 + 28 + 21, or 255 acres. Place 255 
acres under the column of acres. Therefore, the sum is 
255 A. 3 sq. ch. 14 sq. rd. 122 sq. li. Ans. 


( 102 ) Before we can subtract 300 ft. from 20 rd. 2 
yd. 2 ft. and 9 in., we must reduce the 300 ft. to higher 
denominations. 


Since there are 3 feet in 1 yard, in 300 feet there are 
as many yards as 3 is contained times in 300, or 100 yards. 

There are 5^ yards in 1 rod, hence in 100 yards there are as 

1 11 2 

many rods as 5— or — is contained times in 100 = 18— rods. 

2 2 11 




2 

11 ) 200 ( 18-^ rd 
11 

90 

88 


1 11 2 

Since there are 5— or — yards in 1 rod, in — rods there 
2 2 11 


% 


are ^ X 




or one yard, so we find that 300 feet equa’ 


From 


18 rods and 1 yard. The problem now is as follows: 

20 rd. 2 yd. 2 ft. and 9 in. take 18 rd. and 1 yd. 

We place the smaller number under the larger one, so 
that units of the same denomination fall in the same 
column. Beginning with the lowest 
denomination, we see that 0 inches 
from 9 inches leaves 9 inches. Going 
to the next higher denomination, we 
see that 0 feet from 2 feet leaves 
2 feet. Subtracting 1 yard from 2 


rd. yd. 
20 2 
18 1 

T 


in. 

9 

0 




ARITHMETIC. 


51 


yards, we have 1 yard remaining, and 18 rods from 20 rods 
leaves 2 rods. Therefore, the difference is 2 rd. 1 yd. 2 ft, 
9 in. Ans. 


( 103 ) 


A. 

sq. rd. 

sq. yd. 

114 

80 

25 

75 

70 

30 

39 

9 

25^ 


Explanation. —Place the subtrahend under the minuend 
so that like denominations are under each other. Then 
begin at the right with the lowest denomination. We 
can not subtract 30 from 25, so we take one square rod 

(= 30j square yards) from 80 square rods, leaving 79 square 


rods; adding 30- square yards to 25 square yards, we have 
55j square yards; subtracting 30 from 55^ square yards 

leaves 25^ square yards; we now subtract 70 square rods 

from 79 square rods, which leaves 9 square rods; next, vre 
subtract 75 acres from 114 acres, which leaves 39 acres, 
which we place under the column of acres. 


( 104 ) If 10 gal. 2 qt. and 1 pt. of molasses are sold 
from a hogshead at one time, and 26 gal. 3 qt. are sold at 
another time, then the total amount of molasses sold equals 
10 gal. 2 qt. 1 pt. plus 26 gal. 3 qt. 

Since the pint is the lowest denomination, we add the 
pints first, which equal 0 + 1, or 1 pint. We can not reduce 

1 pint to any higher denomina¬ 
tion, so we place it under the 
pint column. The number of 
quarts is 3 + 2, or 5. Since 
there are 4 quarts in 1 gallon, 
in 5 quarts there are as many 
gallons as 4 is contained times in 5, or 1 gallon and 
1 quart remaining. We place the 1 quart under the quart 
column, and reserve the 1 gallon to add to the column of 


gal. 

10 

26 


37 gal. 1 qt. 1 pt. 




ARITHMETIC. 


HP. 


gallons. The number of gallons equals 1 (reserved) -j- ^6 
+ 10, or 37 gallons. 

If 37 gal. 1 qt. and 1 pt. are sold from a hogshead of 
molasses (63 gal.), there remains the difference between 63 
gal. and 37 gal. 1 qt. 1 pt., or 25 gal. 2 qt. and 1 pt. 

63 gal. is the same as 62 gal. 3 qt. 2 pt., since 1 gal. equals 
4 qt. and 1 qt. = 2 pt. 

Beginning with the lowest denomination, 1 pt. from the 
2 pt. 1 pint from 2 pints leaves 1 
2 pint. One quart from 3 quarts 
leaves 2 quarts, and 37 gallons 

- from 62 gallons leaves 25 gallons. 

^ Therefore, there are 25 gal. 2 qt. 


gal. 

62 

37 

25 


qt. 

3 

1 

T" 


and 1 pt. of molasses remaining in the hogshead. Ans. 


( 105 ) If a person were born Jun^ 19, 1850, in order to 
find how old he would be on Aug. 3, 1892, subtract the 
earlier date from the later date. 

On August 3, 7 mo. and 3 da. have elapsed from the begin¬ 
ning of the year, and on June 19, 5 mo. and 19 da. 

Beginning with the lowest denomination, we find that 19 
days can not be taken from 3 days, so we take 1 month from 

which we took equals 30 days, for 
in all cases 30 days are allowed to 
a month. Adding 30 days to the 
3 days, we have 33 days; subtract¬ 
ing 19 days from 33 days, we have 
14 days remaining. Since we bor¬ 
rowed 1 month from the months 
column, we have 7 — 1, or 6 months remaining; subtracting 
5 months from 6 months, we have 1 month remaining. 1850 
from 1892 leaves 42 years. Therefore, he would be 42 years 
1 month and 14 days old. Ans. 

( 106 ) If a note given Aug. 5, 1890, were paid June 3, 
1892, in order to find the length of time it was due, subtract 
the earlier date from the later date. 

Beginning with the lowest denomination, we find that 5 
can not be subtracted from 3, so we take a unit from the next 


7 months. The 1 month 

yr. mo. da. 

1892 7 3 

1850 5 19 

42 1 14 




ARITHMETIC. 


53 


yr- 

1892 

1890 


mo. 

5 

7 


da. 

3 

5 


9 


28 


higher denomination, which is 
months. The 1 month which we 
take equals 30 days. Adding the 30 
days to the 3 days, we have 33 days. 

5 days from 33 days leaves 28 days. 
Since we took 1 month from the months column, only 4 
months remain. 7 months cannot be taken from 4 months, 
so we take 1 year from the years column, which equals 12 
months. 12 months + 4 months = 16 months. 7 months 
from 16 months = 9 months. Since we took 1 year from the 
years column, we have 1892 — 1, or 1891 remaining. 1890 
from 1891 leaves 1 year. Hence, the note ran 1 year 9 
months and 28 days. Ans. 


( 107 ) Write the number, of the year, month, day, hour, 
and minute of the earlier date under the year, month, day, 
hour, and minute of the later date, and subtract. 

22 minutes before 8 o’clock is the same as 38 minutes after 
7 o’clock. 7 o’clock p. m. is 19 hours from the beginning of 
the day, as there are 12 hours in the morning and 7 in the 
afternoon. December is 11 months from the beginning of 
the year. 

10 o’clock A. M. is 10 hours from the beginning of the day. 
July is 6 months from the beginning of the year. The 
minuend would be the later date, or 1,888 years, 11 months, 
11 days, 19 hours, and 38 minutes. 

The subtrahend would be the earlier date, or 1,883 years, 
6 months, 3 days, 10 hours, and 16 minutes. 

Subtracting, we have 


yr. 

mo. 

da. 

hr. 

min. 

1888 

11 

11 

19 

38 

1883 

6 

3 

10 

16 

5 

5 

8 

9 

22 

da. 9 hr. and 22 

min. 

Ans. 


16 minutes subtracted from 38 minutes leaves 22 minutes; 
10 hours from 19 hours leaves 9 hours; 3 days from 11 days 
leaves 8 days; 6 months subtracted from 11 months leaves 
5 months: 1,883 from 1,888 leaves 5 years. 




54 


ARITHMETIC. 


( 108 ) In multiplication of denominate numbers, we 
place the multiplier under the lowest denomination of the 
multiplicand, as 

17 ft. 3 in. 

51 


•879 ft. 9 in. 


and begin at the right to multiply. 51 X 3 = 153 in. As 
there are 12 inches in 1 foot, in 153 in. there are as many 
feet as 12 is contained times in 153, or 12 feet and 9 inches 
remaining. Place the 9 inches under the inches, and reserve 
the 12 feet. 51 X 17 ft. = 867 ft. 867 ft. + ft. (reserved) 
= 879 ft. 

879 feet can be reduced to higher denominations by divi¬ 


ding by 3 feet to find the number of yards, and by 5 — yards 

2 

to find the number of rods. 


3 ) 8 7 9 ft. 9 in. 

5.5 ) 29 3 yd. 

5 3 rd. H yd. 

Then, answer = 53 rd. 1^ yd. 0 ft. 9 in.; or 53 rd. 1 yd. 2 
ft. 3 in. 


( 109 ) qt. 

pt. 

gi- 

3 

1 

3 



4.7* 

1 8.2 qt. 

0 

.1 

or, 1 8 qt. 

0 pt. 

1.7 gi. 

or, 4 gal. 2 qt. 

0 pt. 

1.7 gi. Ans. 


Place the multiplier under the lowest denomination of the 
multiplicand, and proceed to multiply. 4.7 X 3 gi. = 14.1 gi. 
As 4 gi. = 1 pt., there are as many pints in 14.1 gi. as 4 is 
contained times in 14.1 =3.5 pt. and .1 gi. over. Place .1 
under gills and carry the 3.5 pt. forward. 4.7 X 1 pt. = 4.7 
pt.; 4.7 + 3.5 pt. = 8.2 pt. As 2 pt. = 1 qt., there are as 
many quarts in 8.2 pt. as 2 is contained times in 8.2 = 4.1 
qt. and no pints over. Place a cipher under the pints, and 
carry the 4.1 qt. to the next product. 4.7 X 3 qt. = 14.1; 
14.1 + 4.1 = 18.2 qt. The answer now is 18.2 qt. 0 pt. .1 






ARITHMETIC. 


55 


gi. Reducing the fractional part of a quart, we have 18 qt. 
0 pt. 1.7 gi. (.2 qt. = .2 X 8 = 1.6 gi.; 1.6 + .1 gi. = 1.7 gi.). 
Then, we can reduce 18 qt. to gallons (18 4 = 4 gal. and 

2 qt.) = 4 gal. 2 qt. 1.7 gi. Ans. 

The answer may be obtained in another and much easier 
way by reducing all to gills, multiplying by 4.7, and then 
changing back to quarts and pints. Thus, 

3 qt. 1 pt. 3 gi. = 31 gi. 

31 gi. X 4.7 = 145.7 gi. 

4 )145.7 gi. 

2 )36 pt.+1.7gi. 

18 qt. + 0 pt. 

Ans. = 18 qt. 1.7 gi.; 
or, 4 gal. 2 qt. 1.7 gi. 

(IIO) (3 lb. 10 oz. 13 pwt. 12 gr.) X 1.5 = ? 

3 lb. 10 oz. 13 pwt. 12 gr. 

XJ^ 

3 6 oz. 

+19 

4 6 oz. 

X 20 

9 2 0 pwt. 

+ 13 

9 3 3 pwt. 

X 24 
2 2 3 9 2 gr. 

~h I ^ 

2 2 4 0 4 gr. 

22.404 gr. X 1.5 = 33,606 gr. 

24 ) 3 3 6 0 6 gr. 

20) 1 4 0 0 pwt. + 6 gr. 

12 oz. + 0 pwt. 

5 lb. + 10 oz. 


3 qt. 
X 2 pt. 

6 pt. 

+ Ipt. 

7 pt. 
X 4 gi. 
~~^gi. 
+ 3gi. 

31 gi. 











56 


ARITHMETIC. 


Since there are 24 gr. in 1 pwt., in 33,606 gr. there are as 
many pwt. as 24 is contained times in 33,606, or 1,400 pwt. 
and 6 gr. remaining. This gives us the number of grains 
in the answer. We now reduce 1,400 pwt. to higher denom¬ 
inations. Since there are 20 pwt. in 1 oz., in 1,400 pwt. 
there are as many ounces as 20 is contained times in 1,400, 
or 70 oz. and 0 pwt. remaining; therefore, there are 0 pwt. 
in the answer. We reduce 70 oz. to higher denominations. 
Since there are 12 oz. in 1 lb., in 70 oz. there are as 
many pounds as 12 is contained times in 70, or 5 lb. 
and 10 oz. remaining. We can not reduce 5 lb. to any 
higher denominations. Therefore, our answer is 5 lb. 10 oz. 
6 gr. 

Another but more complicated way of working this 
problem is as follows: 


To get rid of the decimal 


lb. oz. pwt. gr. 
3 10 13 12 



pound to ounces. Since 
1 lb. = 12 oz., .5 of a pound 


1.5 


4.5 15 19.5 18 equals .5 lb. X 12 = 6 oz. 

or, 4 21 19 3 0 6 oz. -|-15 oz. = 21 oz. We 

or, 5 10 0 6 Ans. now have 4 lb. 21 oz. 19.5 pwt. 


and 18 gr., but we still have a 


decimal in the column of pwt., so we reduce .5 pwt. to grains 
to get rid of it. Since 1 pwt. = 24 gr., .5 pwt. = .5 pwt 
X 24 12 gr. 12 gr. + 18 gr. = 30 gr. We now have 4 
lb. 21 oz. 19 pwt. and 30 gr. Since there are 24gr. in 1 pwt., 
in 30 gr. there is 1 pwt. and 6 gr. remaining. Place 6 gr. 
under the column of grains and add 1 pwt. to the pwt. 
column. Adding 1 pwt., we have 19 -j- 1 = 20 pwt. Since 
there are 20 pwt. in 1 oz., we have 1 oz. and 0 pwt. remain¬ 
ing. Write the 0 pwt. under the pwt. column, and reserve 
the 1 oz. to the oz. column. 21 oz. 1 oz. = 22 oz. Since 
there are 12 oz. in 1 lb., in 22 oz. there is 1 lb. and 10 oz. 
remaining. Write the 10 oz. under the ounce column, and 
reserve the 1 lb. to add to the lb. column. 4 lb. -f 1 lb- 
(reserved) = 5 lb. Hence, the answer equals 5 lb. 10 oz. 
6 gr. 



ARITHMETIC. 


57 


(111) If each barrel of apples contains 2 bu. 3 pk. and 
6 qt., then 9 bbl. will contain 9 X (2 bu. 3 pk. 6 qt.). 

We write the multiplier under the lowest denomination of 
the multiplicand, which is quarts in this problem. 9 times 
6 qt. equals 54 qt. There are 8 qt. in 1 
bu. pk. qt. pk., and in 54 qt. there are as many pecks 
^ 3 6 as 8 is contained times in 54 , or 6 pk. and 

9 6 qt. We write the 6 qt. under the col- 

18 2 7 5 4 umn of quarts, and reserve the 6 pk. to 

or, 2 6 1 6 the product of the pecks. 9 times 

3 pk. equals 27 pk. ; 27 pk. plus the 6 pk. 
reserved equals 33 pk. Since there are 4 pk. in 1 bu., in 33 
pk. there are as many bushels as 4 is contained times in 33, 
or 8 bu. and 1 pk. remaining. We write the 1 pk. under 
the column of pecks, and reserve the 8 bu. for the product 
of the bushels. 9 times 2 bu. plus the 8 bu. reserved equals 
26 bu. Therefore, we find that 9 bbl. contain 26 bu. 1 pk. 
6 qt. of apples. Ans. 

(112) (7 T. 15 cwt. 10.5 lb.) X 1.7= ? When the mul¬ 

tiplier is a decimal, instead of multiplying the denominate 
numbers as in the case when the multiplier is a whole num¬ 
ber, it is much easier to reduce the denominate numbers to 
the lowest denomination given; then, multiply that result 
by the decimal, and, lastly, reduce the product to higher 
denominations. Although the correct answer can be ob¬ 
tained by working examples involving decimals in the 
manner as in the last example, it is much more complicated 
than this method. 7 T. 15 cwt. 10.5 lb. 

X 

14 0 cwt. 

15 

15 5 cwt. 

X 100 

15 5 0 0 lb. 

10.5 

15510.5 lb. 

15,510.5 lb. X 1.7 = 26,367.85 lb. 

H,G. IV.—5 


I 





58 


ARITHMETIC. 


There are 100 lb. in 1 cwt., and in 26,367.85 lb. there are 
as many cwt. as 100 is contained times in 26,367.85, which 

equals 263 cwt. and 67.85 lb. 

100) 2 6 3 6 7.8 5 lb. remaining. Since we have 

20) 2 6 3 cwt. + 67.85 lb. the number of pounds for 

1 3 T. + 3 cwt. our answer, we reduce 263 

cwt. to higher denominations 
There are 20 cwt. in 1 ton, and in 263 cwt. there are as 
many tons as 20 is contained times in 263, or 13 tons and 3 
cwt. remaining. Since we cannot reduce 13 tons any 
higher, our answer is 13 T. 3 cwt. 67.85 lb. Or, since .85 
lb. = .85 lb. X 16 = 13.6 oz., the answer may be written 
13 T. 3 cwt. 67 lb. 13.6 oz. 

( 113 ) 7 ) 358 A. 57 sq. rd. 6 sq. yd. 2 sq. ft. 

51 A. 31 sq. rd. 0 sq. yd. 8 sq. ft. Ans. 

We begin with the highest denomination, and divide each 
term in succession by 7. 

7 is contained in 358 A. 51 times and 1 A. remaining. 
We write the 51 A. under the 358 A. and reduce the remain¬ 
ing 1 A. to square rods = 160 sq. rd.; 160 sq. rd. +'the 57 
sq. rd. in the dividend = 217 sq. rd. 7 is contained in 217 
sq. rd. 31 times and 0 sq. rd. remaining. 7 is not contained 
in 6 sq. yd., so we write 0 under the sq. yd. and reduce 
6 sq. yd. to square feet. 9 sq. ft. X 6 = 54 sq. ft. 54 sq. ft. 
+ 2 sq. ft. in the dividend = 56 sq. ft. 7 is contained in 
56 sq. ft. 8 times. We write 8 under the 2 sq. ft. in the 
dividend. 

( 114 ) 12 ) 282 bu. 3 pk. 1 qt. 1 pt. 

23 bu. 2 pk. 2 qt. ^ pt. Ans. 

12 is contained in 282 bu. 23 times and 6 bu. remaining. 

We write 23 bu. under the 282 bu. in the dividend, and 

reduce the remaining 6 bu. to pecks = 24 pk. -f the 3 pk. in 
the dividend = 27 pk. 12 is contained in 27 pk. 2 times and 
3 pk. remaining. We write 2 pk. under the 3 pk. in the 
dividend, and reduce the remaining 3 pk. to quarts. 3 pk. 
= 24 qt.; 24 qt. + the 1 qt. in the dividend = 25 qt. 12 is 
contained in 25 qt. 2 times and 1 qt. remaining. We write 






ARITHMETIC. 


59 


2 qt. under the 1 qt. in the dividend, and reduce 1 qt. to 
pints = 2 pt. + the 1 pt. in the dividend = 3 pt. 3 -t- 12 = 



pt. 


( 115 ) We must first reduce 23 miles to feet before we 
can divide by 30 feet. 1 mi. contains 5,280 ft.; hence, 23 
mi. contain 5,280 X 23 = 121,440 ft. 

121,440 ft. 30 ft. = 4,048 rails for 1 side of the track. 

The number of rails for 2 sides of the track = 2 X 4,048, 
or 8,096 rails. Ans. 


( 116 ) In this case where both dividend and divisor are 
compound, reduce each to the lowest denomination men¬ 
tioned in either and then divide as in simple numbers. 


1 bu. 1 pk. 7 qt. 

3 5 6 bu. 3 pk. 5 qt. 

X 4 

X 4 

4 pk. 

1424pk. 

±1 

+ 3 

5 pk. 

1427pk. 

X 8 

X 8 

40 qt. 

1141 6 qt. 

±1 

+ 5 

4 7 qt. 

11 4 2 1 qt. 

47)11421(243 


94 

11,421 qt.-7- 47 qt. = 243 boxes 

2^ 

Ans. 

188 


141 


141 


( 117 ) We must first reduce 16 square miles to acres. 

In 1 sq. mi. there are 640 A., and in 16 sq. mi. there are 

1 X 640 A. = 10,240 A. 


62 ) 10240 A. 


] 65 A. 25sq.rd. 

24sq.yd. 3sq.ft. 804-sq. in. Ans. 









60 


ARITHMETIC. 


62 is contained in 10,240 A. 165 times and 10 A. remain^ 
ing. We write 165 A. under the 10,240 A. in the dividend 
and reduce 10 A. to sq. rd. In 1 A. there are 160 sq. rd., 
and in 10 A. there are 10 X 160 = 1,600 sq. rd. 62 is con¬ 
tained in 1,600 sq. rd. 25 times and 50 sq. rd. remaining. 
We write 25 sq. rd. in the quotient and reduce 50 sq. rd. to 

sq. yd. In 1 sq. rd. there are 00^ sq. yd., and in 50 sq. rd. 


there are 50 times 30y sq. yd. = 1,512^ sq. yd. 62 is con- 
4 2 

tained in 1,512^ sq. yd. 24 times and 24— sq. yd. remaining. 
2 2 

In 1 sq. yd. there are 9 sq. ft., and in 24— sq. yd. there are 

2 

24^ X 9 = 220^ sq. ft. 62 is contained in 220^ sq.ft. 3 times 
2 2 2 

and 34^ sq. ft. remaining. We write 3 sq. ft. in the quo- 
2 


tient and reduce 34-^ sq. ft. to sq. in. In 1 sq. ft. there are 
2 

144 sq. in., and in 34-i sq. ft. there are 34-^ X 144 = 4,968 
2 2 

sq. in. 62 is contained in 4,968 sq. in. 80 times and 8 sq. in. 
remaining. 

We write 80 sq. in. in the quotient. 

It should be borne in mind that it is only for the purpose 
of illustrating the method that this problem is carried out 
to square inches. It is not customary to reduce any lower 
than square rods in calculating the area of a farm. 


ARITHMETIC. 

(SECTION 5.) 
(QUESTIONS 118-133.) 


(118) To square s number, we must multiply the num¬ 
ber by itself once, that is, use the number twice as a factor 
Thus, the second power of 108 is 108 X 108 = 11,664. Ans 

108 

108 

864 

1080 

11664 

( 119 ) 9^= 9 X 9 X 9 X 9 X 9 = 59,049. Ans. 

9 

9 

9 

7^9 

9 

6 5 6 1 
9 

5 9 049 

( 120 ) (a) .0133® = .0133 X. 0133 X. 0133 = .000002352637. 

Ans. 


For notice of copyright, see page immediately following the title page. 







G2 


ARITHMETIC. 


Since there are four decimal places in the multiplicand and 


4 in the multiplier, we 

.0133 
.0 133 


3 9 9 
399 
133 


.0 0017689 
.0 133 

5 30 6 7 
5 3 0 6 7 
1 7 6 8 9 

.0 00002352637 


must point off 4 -|- 4 = 8 decimal 
places in the product; but as 
there are only 5 figures in the 
product, we prefix three ciphers 
to form the eight necessary deci- 
mal places in the first product. 

Since there are 8 decimal places 
in the multiplicand and 4 in the 
multiplier, we must point off 8 
-f- 4 = 12 decimal places in the 
product; but as there are only 
7 figures in the product, we prefix 
5 ciphers to make the 12 neces¬ 
sary decimal places in the final 
product. 


(121) Evolution is the reverse of involution. In invo¬ 
lution we find the power of a number by multiplying the 
number by itself one or more times, while in evolution we 
find the niimber or root which was multiplied by. itself one or 
more times to make the power. 


(122) 4/^ = \ The root is evidently 9 plus an intermi¬ 
nable decimal. Trying 9 for one factor, the other is 90 9 


= 10, and the first approximation is^-i-^ = 9.5. 


90 -^ 


= 9.473+, and the second approximation is 


9.486+, or 9.49 to three figures. Using 9.49 for one fac¬ 
tor, the other is 909.49 = 9.48366+, and the third 
. 9.49 + 9.48366 

approximation is--h_--= 9.48683 or 9.4868+ to 

2 


five figures. Ans. 

This solution may be shortened by using the table and 
applying the method described in Art. 281 to find the first 
three significant figures of the root. Referring to the table, 
the first two significant figures of the root are 9.4; the first 
difference is 90.25 — 88.36 = 1.89; the second difference is 









ARITHMETIC. 


03 


90 — 88.36 = 1.64; 1.64-^ 1.89 = .86-|-* Hence, the first 
three figures are 9.48. 


(1 23) To find any power of a mixed number, first reduce 
it to an improper fraction, and then multiply the numerators 
together for the numerator of the answer, and multiply the 
denominators together for the denominator of the answer. 

/o3\* _ 15 15 15 _ 15 X 15 X 15 _ 3.375 _ ,,^47 

4^4 4 4X4X4 64 “ 64 

= 52.734375. Ans. 


^3 _ 3 X 4 -f 3 
"^4 4 

15 

15 

15 

225 

15 

1125 

225 

33 75 


12 + 3 _ 15 
4 “4* 

47 

64)3375(52^ 

3 2 0 

T75 

128 

"Tr 


64)4 7.0 00000(.7 34375 
448 

Tio 

192 

~so 

256 

Tio 

192 

Tio 

448 

~20 

320 








64 


ARITHMETIC. 


Since six ciphers were annexed to the dividend, six decimal 
places must be pointed off in the quotient. 


( 124 ) 4 / 92,416 = ? Pointing off into periods, the result 

is 92'416. Since 92 lies between 4® = 64 and 5® = 125, the 
root is 44 -. Trying 4 for one of the two equal factors, the 
third factor is 92 ^ 4“* = 92 ^ 16 = 5.75. Trying 5 for one 
of the two equal factors, the third factor is 92 -h 5® = 92 -h 25 
= 3.68. Difference between 4 and 5.75 is 1.75, and between 
5 and 3.68 is 1.32; hence, use 5, the first approximation 


being 


2 X 5 + 3.68 


= 4.56, or 4.6 to two figures. 


Using 46 for one of the two equal factors, the third factor 
is 92416 -^46® = 92416 -^ 2116 = 43.67+, and the second 

approximation is- ^ -- = 45.22+, or 45.2 to three 

O 

figures. 

Using 45.2 for one of the two equal factors, the third 

factor is 92416 45.2® = 92416 ^ 2043.04 = 45.2345+, and 

^ .2x 45.2 + 45.2345+ 

the third approximation is----— = 45.2115 + , 

O 

or 45.212— to five figures. Ans. 

The first three significant figures may also be found by the 
aid of the table and the method described in Art. 282. 

In order to obtain two figures of the root from the table, 
we place a decimal point between the first and second signifi¬ 
cant periods; the result is 92.416. Referring to the table, the 
first two figures of the root are 4.5; the first difference is 
97.336 — 91.125 = 6.211; the second difference is 92.416 
- 91.125 = 1.29]jJ.291-^ 6.211 =.20+. Therefore, l/O^Hie 
= 4.52 and V^92,416 = 45.2 to three significant figures. 


( 125 ) 4 / 502,681 = ? Pointing off into periods, we have 

50'26'81. The first figure of the root is evidently 7, since 7® 
= 49 and 8® = 64. The two factors then are 7, and 50 7 

= 7.14+. The first approximation is — 7-14:+ _ 7 . 07 +, 

or 7.1 to two figures. To find the second approximation, we 
use the first two periods and drop the decimal point + f^he first 










ARITHMETIC. 


65 


approximation. One factor is then 71 and the other 5026 
71 = 70.78+. The second approximation is therefore 

r , I 70 7Q ' 

--—^— = 70.89, or 70.9 to three figures. Using 709 for 


one factor, the other is 502681 709 = 709. Hence, the 

number is a perfect power and the root is 709. Ans. 

Using the table and placing a decimal point between the 
first and second periods so that we may obtain two figures of 
the root from the table, the number becomes 4/50.2681 or to 
four figures, 4 / 5 O. 27 . Referring to the table, the first two 
figures of the root are 7.0; the first difference is 50.41 — 49.00 
= 1.41; the second difference is 50.27 — 49.00 = 1.27; 1.27 

1.41 = .9+. Therefore, 4 / 502 , 68 ! = 709 to three figures. 


( 126 ) 

( 127 ) 



V¥i 


4’ = 4 X 4 X 4 = 64. 


Ans. 


^ = 3 . 

4* - Vs = 64 - 3 = 62. Ans. 

(128) Sinceg =.375,= V-375. Moving the deci- 

mal point three places to the right, the number becomes 375. 
Since 375 lies between 7® = 343 and 8® = 512, the root is 7+. 
Trying 7 for one of the two equal factors, the third factor is 
375 -V 7“ = 375 49 = 7.65-1-, and the first approximation is 

^ ^ ^ or 7.2 to two figures. As the differ- 

o 


ence between the equal and unequal factors is very slight, it 
is not necessary to try 8. 

Using 7.2 for one of the two equal factors, the third 
factor is 375 U- 7.3” = 375 51.84 = 7.333+, and the second 

approximation is- - -= 7.211+, or 7.21 to three 

O 

figures. 

Using 7.21 for one of the two equal factors, the third 
factor is 375 -f-7.21® = 375 ^ 51.9841 = 7.21374-f, and the 

third approximation is - - -— = 7.21124+, or 

O 









66 


ARITHMETIC. 


7.2112+ to five figures. Since the number is entirely deci 
mal, the root is wholly decimal; hence, locating the decimal 
point, the-^.375 = .72112+. Ans. 

By aid of the table, the first three figures are determined 
as follows: Move the decimal point three places to the right 
so it will fall between the first period 375 and the cipher 
period that follows. 

Referring to the table, the first two figures of the root are 
7.2; the first difference is 389. 17 - ^7'.248 = 15.769; the 
second difference is 375.000 - 373.248 = 1.752; 1.752 
15.769 = .11+, or .1 to one figure. Hence, the first three 
figures are 7.21, or the |/. 375 = .721. The fourth and fifth 
figures are then determined as previously indicated. 

(129) |/:M64=? 

Moving the decimal point two places to the right, so that 
the first period may be integral, the result is 33.64. The first 
two factors are evidently 5 and 33 -J- 5 = 6 . 6 , and the first 

approximation is — = 5. 8 . 33.64 5 .8 = 5. 8 . Hence, 

the given number is a perfect power, and as it is wholly 
decimal, 4 /. 3364 = .58. Ans. 

( 130 ) |/3.1416 = ? 

Pointing off, we obtain 3.14'16. The first twft significant 
figures are 3.1. It is evident that the first figure of the root 
is 1 , since P = 1 and 2" = 4. Using 1 as one factor, the other 

14- 3 1 

is 3.1 1 = 3.1, and the first approximation is —h—^ = 2.05. 

2 

Had 2 been used as one factor, the other would have been 
3.1 -J- 2 = 1.55, and the first approximati n would have been 
2 + 1 55 

-^— = 1.V7+, or 1.8 to two figures. In the first case, 

the difference between the two factors is 3.1 — 1 = 2 . 1 ; in 
the second case, the difference is 2 — 1.55 = . 45 . As the 
factors are more nearly equal in the second case than in the 
first, it is evident that 1.8 is more nearly equal to the correct 
ralue of the root than 2.05 is; hence, 1.8 will be used for the 
first approximation. 








ARITHMETIC. 


67 


For the second approximation, use the first two periods 
and 1.8 for one factor, the other factor is 3.14 4- 1.8 = 1.74+ ; 


hence, the second approximation = 


1.8 + 1.74 
2 


= 1.77. 


Using 


1.77 for one factor, the other is 3.1416 1.77 = 1.77491+. 

1 77 4_ 1 77491 

The third approximation is - = 1.772455, or 

1.7725— to five figures. Ans, 

Using the table to find the first three figures, the first two 
figures of the root are 1.7; the first difference is 3.24 — 2.89 
= .35; the second difference is 3.14 — 2.89 = .25; .25-1-.35 
= .71+. Therefore, +3.1416 = 1.77 to three significant 
figures. 


( 131 ) Since some number multiplied by itself equals 
114.9184, then the number is |/114.9184. Pointing off into 
periods and placing the decimal point between the first and 
second periods, we have 1.14'91'84. Considering the first 
two figures, it is evident that the first figure of the root is 1. 
Using 1 as one factor, the other is 1.1 -i- 1 = 1.1, and the first 


approximation is 


1 + 1.1 

2 


1.05, or 1.1 to two figures. Using 


the first three figures and 1.1 for one factor, the other factor 
is 1.151.1 = 1.045+, and the second approximation is 

+ l-04u _ 2 072+, or 1.07 to three figures. Using 1.07 

for one factor, the other is 1.149184-^ 1.07 = 1.074003+, 

and the third approximation is — — = 1.072001+, 

or 1.0720 to five figures. Noticing that the square of the last 
significant figure is 2’ = 4, which corresponds to the last figure 
of the given number, and that the fifth and sixth figures of the 
third approximation are ciphers, we suspect that the given 
number is a perfect power. We find such to be the case on 
squaring 1.072. Since there are two periods in the integral 
part of the number, there are two figures in the integral part 
of the root, and 4/114.9184 = 10.72. Ans. 

Using the table to find the first three figures of the root, 
the first two figures of the root are 1.0; the first difference 










08 


ARITHMETIC. 


is 1.21 — 1.00 = .21; the second difference is 1.15—100 
= .15; .15-T-.21 = .7. Therefore, the first three figures 
are 10.7. 


(132) 4^3,486,784 = ? 

Pointing off into periods, we have 3'48'67'84. Placing a 
decimal point between the first and second periods and 
using the first two figures, we obtain 3.5. Considering 2 as 
one factor, the other is 3.5 2 = 1.75, and the first approxi¬ 


mation is 


2 + 1.75 
2 


= 1.87+, or 1.9 to two figures. 


For the 


second approximation, we use the first two periods and 19 for 
one factor, the other factor being 349 19 = 18.37+. We 

used 349 instead of 348 because the fourth figure was 6, and 
the number correct to three figures is 349. The second 
19 +18 37 

approximation is --——= 18.68+, or 18.7 to three 

figures. 34868 187 = 186.459+ ; + 186.459 _ 739+, 

or 1867.3— to five figures. Ans. 

In order to obtain three figures from the table, we place 
the decimal point between the first and second periods, 
and use the first two periods only; that is, we find the value 
of 4/3.49. Referring to the table, the first two figures of the 
root are 1.8; the first difference is 3.61 — 3.24 = .37; the 
second difference is 3.49 — 3.24 = .25; .25 -f- .37 = .67+, 
or .7 to one figure. Therefore, 4/3,486,784 = 187 to three 
significant figures. 

The fourth and fifth figures may be found as previously 
indicated. 


(133) 4/. 00041209 = ? 

Pointing off into periods, the result is .00'04T2'09. Placing 
the decimal point between the first and second periods of the 
significant part of the number, we obtain 4.1 for the first two 
figures. The first factor is evidently 2 and the second factor 

4.1 -i- 2 2.05. The first approximation is —= 2.02+, 

or 2.0 to two figures. Using 20 for one factor, the other is 









ARITHMETIC. 


69 


412 -f- 20 = 20.6, and the second approximation is ^ 

= 30.3. ' 

Using 203 for one factor, the other is 41209 203 •— 203. 

Hence the number is a perfect power and the significant 
figures of the root are 203. There being one full cipher period 
following the decimal.point, the root is .0203. Ans. 

Using the table, we find that the first two figures of the 
root are 2.0; the first difference is 4.41 — 4.00 =.41; the sec¬ 
ond difference is 4.12 — 4.00 = .12; .12-j-.41 = .3, nearly. 
Therefore, |/. 00041209 = .0203 to three significant figures. 








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ARITHMETIC 

(SECTION 6.) 
(QUESTIONS 143-167.) 


( 143 ) 11.7 : 13::20 : x. The product of the means 

11.7^z= 13x20 equals the product of the 

= 260 extremes. 

_ ^ 

11.7 ) 260.000 ( 22.22+ Ans. 

234 

2T0 

234 

260 

234 

260 

234 

Te 

( 144 ) {a) 20 + 7 : 10 + 8:: 3: .r. 

27 : 18::3 : 

27 .^r= 18 X 3 
27.r = 54 

. = -^ = 2 . Ans. 

(^) 12’ : 100’4 : j:. 

144 : 10,000:: 4 : x 
144 x= 10,000 X 4 
144 x= 40,000 

For notice of copyright, see page immediately following the title page. 





ARITHMETIC. 


TZ 


_ 40,000 

144 ) 40000.0 ( 277.7+ Ans. 

288 

1120 

1008 

1120 

1008 

1120 

1008 

112 

(145) (a) ~ is equivalent to 4 : ;ir:: 7 : 21. The 

product of the means equals the product of the extremes. 
Hence, 

7.:r==r4x2X 
7;r = 84 
84 

= — or 12. Ans. 

{d) In like manner, 

3 

^ = jg is equivalent to ;r : 24 :: 8 : 16 
16 4r= 24 X 8 
16;r = 192 


^ = -^g-=13. 


Ans. 




2 X 

^ is equivalent to 2 : 10:: .sr : 100. 


10.r = 2 X 100 
10;r = 200 
200 
10 


= 'ZO. Ans. 


. 15 60. . - 

(«) — = — IS equivalent to 

15 : 45:: 60 : ;ir. 

15;r = 45 X 60 
15;tr = 2,700 
2,700 
15 


X = ■ 


= 18a 


Ans. 


/ X 10 X . . . 

= equivalent to 

10 : 150 :: .r : 600. 

150 = 10 X 600 

150 ;i: = 6,000 

C>000 ^ 

^ = -j^-40. Ans. 







ARITHMETIC. 


73 


(146) ;r : 5 :: 27 : 13.5. (147) 

45 : 60 :: .r : 24 

5 

60.r= 45X24 

12.5)135.0(loi Ans. 

60^"= 1,080 

125 ® 

1,080 ^ 

, = 18. Ans. 

100 4 

60 

1 2 5 “ 5 


(148) .r : 35 :: 4 : 7. (149) 

9 : -r :: 6 : 24. 

7 .r = 35 X 4 

6 = 9 X 24 

7x= 140 

6.;r = 216 

140 

216 

.r = — = 20. Ans. 

;r = ^ = 36. Ans. 
6 

(ISO) 


^1,000 : -^1,331 :: 27 : ;r. 

^1,000 = 10. 

10 : 11 ::27 : .r. 

^1,331 = 11. 

10jr=:297. 1 

1 1'331(11 

^97 ^ 1 

2 1 

r = —= 29.7. _ 


10 9 

Ans. ^ 

300 331 


31 331 

3 0 

331 


1 

31 

(151) 64 : 81 = 2V : 

Extracting the square root of each term of any proportion 
does not change its value, so we find that : 4/^ = 
is the same as 
8 : 9 


Ans. 


21 : 

Sx= 189 

23.625. 

(152) 7 + 8 : 7 = 30 : .tr is equivalent to 

15 : 7 = 30 : ;r. 

15;r = 7 X 30 
15 x= 210 
210 
^“15 


= 14. Ans. 


H. Q. IV.—6 










74 


ARITHMETIC. 


(153) 2 ft. 5 in. = 29 in.; 2 ft. 7 in. = 31 in. Stating 
as a direct proportion, 29 : 31 = 2,480 : .ir. Now, it is easy 
to see that will be greater than 2,480. But x should be 
less than 2,480, since, when a man lengthens his steps, the 
number of steps required for the same distance is less; 
hence, the proportion is an inverse one, and 

29 : 31 = .r : 2,480, 
or, 31.^^ = 71,920; 

whence, .;r = 71,92031 = 2,320 steps. Ans. 

(154) This is evidently a direct proportion. 1 hr. 
36 min. = 96 min.; 15 hr. = 900 min. Hence, 

96 : 900 = 12 : X, 
or, 96.^= 10,800; 

whence, x = 10,800 96 = 112.5 mi. Ans. 

(155) This is also a direct proportion; hence, 

27.63 : 29.4 = .76 : .r, 
or, 27.63.^r = 29.4 X .76 = 22.344; 
whence, .r = 22.344 -t- 27.63 = .808 + lb. Ans. 

(156) 2 gal. 3 qt. 1 pt. =23 pt.; 5 gal. 3 qt. =46 pt. 
Hence, 

23 : 46 = 5 : .r, 
or, 23 = 46 X 5 = 230; 

whence, x = 230 23 = 10 days. Ans. 

(157) Stating as a direct proportion, and squaring the 
distances, as directed by the statement of the example, 
6’ : 12’* = 24 : X. Inverting the second couplet, since this 
is an inverse proportion, 

6’ : 12" = : 24. 

Dividing both terms of the first couplet (see Art. 310) 
by 6 

1" : 2" = ;ir : 24; or 1 : 4 = .r : 24; 

whence, 4 r = 24, or = 6 degrees. Ans. 


ARITHMETIC. 


75 


( 158 ) Taking the dimensions as the causes, 


n 

2 


=n 


whence, 2 jt: = 75, or, x = $37.50. 

Ans. 


or 


( 159 ) ^ hr. — 120 min.; 14 hr. 28 min. = 868 min. 
Hence, 120 : 868 = 100 : ;ir, 

or, 120.r = 86,800; 
whence, .r = 723^ gal. Ans. 

( 160 ) Taking the dimensions as the causes. 


X, whence, 2 ;r = 17 X 57 = 969, 
or, .r = 484i bbl. Ans. 


(161) 8 hr. 40 min. = 520 min. Hence, 

444 : 1,060 = 520 : .^r, 

130 

,= i«^ = ffiSI2=l,2,I.44 + ™, = 20 hr. 

^ 41.44+min. Ans. 



(Z 

<2? 


2 


;2 

17 




(162) 1 min. = 60 sec. Hence,' 

Si : 60= 6,160 : .;r, 
_ 60 X 6,160 


67,200 ft. Ans. 


( 163 ) Writing the statement as a direct proportion, 
8 : 10 = 5 : ;r, it is easy to see that x will be greater than 5; 
but, it should be smaller, since by working longer hours, 
fewer men will be required to do the same work. Hence, 
the proportion is inverse. Inverting the second couplet, 


8:10 = .r ; 5, 
4 


or, 


n 




= 4 men. 


Ans, 










76 


ARITHMETIC. 


(164) Taking the times as the causes, 
P 14 


2 


3 


); whence, 3;r = 2 X 14 = 28, orx= 9^ hr. 

Ans. 


(165) Taking the horsepowers as the effects, we have 
for the known causes in example 4, Art. 349, 14®, 500, and 
48, and for the known effect 112 horsepower. Hence, 

9 


14“ 

500 

48 


30“ 

660= 112 
42 


m 


23 






whence, ;r=9x 22 X3 = 594 horsepower. Ans. 

(166) First find the volume of the cylinder in cubic 
inches, as in the example, Art. 345. The volume, multi¬ 
plied by the weight of one cubic inch (.2011b.), will evidently 
be the weight of the cylinder. Hence, 


10 “ 

20 


= 1,570.8 


100 


or 




144 

3 = 1.570.8 




whence, .r = X 6,785.856 cu. in. Therefore 

weight of cylinder = 6,785.856 X .261 = 1,771.11 lb. Ans 

(167) Referring to the example in Art. 348, 

5 

40 

100 
x^ or ^(30 


15 

20 * 

10 


18“ = 187 
12 


324 = 187 
4 

. 

whence, = ^84.7 lb. Ans. 


















Mensuration and Use of Letters 
IN Formulas. 

(QUESTIONS 142-213.) 


(142) Substituting for B, and i their values. 

r- 108_g . 

“ 10 + 3.5 ~ 13.5“ • 

A line between two numbers signifies that the one above 
the line, or numerator, is to be divided by the one below 
the line, or denominator. 

(143) Substituting for A and x their values, 

Ah-\-D 200)-h 120 _ 1,000+ 120 _ 1,120 _ 

2.r + 6 (2 X 12) + 6 “ 24+6 “ 30 "" 

37i + = 37i + 120 = 157^. Ans. . 

When there is no sign between the letters, multiplication 
is understood. 


( 144 ) Substituting for B, //, Ay Xy and i their values, 
__ 3.246 xBxh _ 3.246 X 10 X 200 _ 6,492 _ 

“ Ax-^rh “ (5 X 12) + 200 “ 260 “ 

A i — B (5X3.5)-10 7.5 

6,4924^ = 6,492 X-^ = 187.269+. Ans. 

7. o /IvO 


( 145 ) Substituting for Ay By /, and B their values. 




5 X 120 


zB-\- 1-5 


(3.5 X 10) + 1.5 ^ 36.5 


4/16:4383 = 4.05 +. Ans. 


The square root sign extends over both numerator and 
denominator, thus indicating that the square root of the 
entire fraction is to be extracted. 


For notice of the copyright, see page immediately following the title page. 


















94 


MENSURATION. 


(146) Substituting for ;r, //, and A their values, 


Bx 


.00018 A (A"-x) 




(147) 


120 


.030 X (26 - 12) 




10 X 12 


.00018 X 200 X (5" - 
120 _ 

~ .468 


12 ) 


.036 X 13 
f ^4i + = 6.35 +. Ans. 
Substituting for h, and A their values, 
10X80 


\^{h—DY _ 10 ( 200 - 120 ) 

' ^ DA-A ~ >^ 120+“5 ~ 


^ = 12,800 

^ Ans. 


(148) Substituting for B, A, and D their values, 

5 )'“ - ^120 +"5 _ 

- (1 + 120) - 


.r 


_ (B- Ay - + A _ (10 


A^-(L + £>) 

5^ — '25—5 


20 ^ 
= ^ = 5. 


Ans. 



125 - 121 4 

Substituting for A, B^ and h their values, 
A A’'_ ^5 X 10 






5 X 100 


500 

10 


f5 X 200 f 1,000 
4/50 = 7.071+. Ans. 

(150) Substituting for A, D, and B their values. 


T= 


[^490-f 


(ML~\ 
D' J 


h-y^{A^-By 


200 + j-^(5“-10)’ 


. / 25(490 + 40 0) _ ./ 25 X 890 _ ./22,250 

' 200 + (tV X 225) “ 200 + 22.6“' 222.5 “ 

I/TOO = 10. Ans. 

_ (151) When one straight line meets another straight 
line, two angles are formed which together equal 180°. 
Hence, if one of the angles = 152° 3', the other angle 
180° - 152° 3', or 

180° = 179° 60' 
subtracting, 152° 3' 


27° 57'. Ans. 










































MENSURATION. 


95 


(152) There are GO seconds in one minute and 60 min¬ 
utes in one degree; therefore, 140° = 140 X 60 = 8,400 min¬ 
utes; 8,400'+ 17' = 8,417'; 8,417' = 8,417 X 60 = 505,020 
seconds, and 505,020" + 10" = 505,030". Ans. 

(153) See Arts. 359 and 360. 

(154) {a) 240 H- 60 = 4, the number of degrees. Ans. 
{b) 240 X 60 = 14,400, the number of seconds. 

Ans. 

(155) See Arts. 355-357. 

(156) See Art. 369. If the rectangle has the same 
base and altitude, it would have the same area. 

(157) No, since the sum of the three shorter sides is 
not greater than the fourth side. 

(158) Since the area is to be found in square inches, 
the 2^ feet must be reduced to inches. 2J ft. = 30 in. Area 
= 30 X Hi — 345 sq. in. Ans. 

(159) Since there are 144 sq. in. in 1 sq. ft., the area of 

345 

the zinc in the last example = = 2.396 sq. ft. The weight 

5 25 

per square foot = —= 2.19 + lb. Ans. 

Z. 090 

(160) It will take 1^ boards to reach lengthways of the 
room. Since the room is 15 feet wide, and each board is 5 

5 

inches wide, it will take 15 = 36 boards, laid side by side, 

1/V 

to extend across the width of the room. Hence, number of 
boards required = 36 X li = 54. Ans. 

(161) By the rule for the area of a trapezoid, area of the 
9 4-6 

land = —— X 16 = 120 square rods. Since.there are 160 
2 

4 • -.OA ^ 120 3 . 

square rods in one acre, 120 square rods = ~ 4 

acre. Ans. 

(162) The total area of the floor of the station 
= 55 X 58 ft. = 3,190 sq. ft. — 25 X 26 ft. = 650 sq. ft., the 
area represented by the lower right-hand corner of the 




96 


MENSURATION. 


figure. Hence, total area of floor = 3,190 — 650 = 2,540 
sq. ft. 

From this we have to deduct the following areas: 

2 boilers = 2x8x19 = 304. sq. ft. 

Feedpump = 2^X5 = 12.5 sq.ft. 

3 engines = 2 X 4^ X 10 = 90 sq. ft. 

2 dynamos — 2 X X 71.5 sq. ft. 

If) V ^ 5 

Switchboard = • ^ = 2.92 sq. ft. 

480.92 sq.ft. 

The unoccupied floor space, therefore, = 

2,540 - 480.92 = 2,059.08 sq. ft. Ans. 

( 163 ) The length of the walk, neglecting the corners, 

that is, the distance 
a b-\-b c-\-c d-\-da. Fig. 
1, = 2 X 528 + 2 X 352 = 
1,760 ft. The width = 10 
ft.; area = 1,760 X 10 = 
17,600 sq. ft. The area of 
the four corners = 4 X 10’ 
= 400 sq. ft. Total area 
of walk = 17,600 + 400 = 

18,000 sq. ft. Hence the area in sq. yds. — 18>000 _ 2,000. 
Ans. ^ 

( 164 ) The area of the sides of the room = 2 X 20 X 90 
= 3,600 sq. ft.; area of ends = 2 X 20 X 50 = 2,000 sq. ft. 
Total area of walls = 3,600 + 2,000 = 5,600 sq. ft. 

From this is to be deducted the following areas; 

4 doors = 4 X 5J X 10 = 220 sq. ft. 

14 windows = 14 X 5 X 11 = 770 sq. ft. 

Length of baseboard, deducting the width of the 4 doors 
= 2 X 90 + 2 X 50 - 4 X 5^ = 180 + 100 - 22 = 258 feet; 

width of baseboard = 9 inches, or f of a foot; area of base- 

3 

board = 358 X= 193.5 sq. ft. Hence, we have to deduct 
330 + 770 + 193.5 = 1,183.5 sq. ft 











MENSURATION. 


97 


Number of sq. ft. of plastering = 5, GOO — 1,183.5 = 4,416.5 
sq. ft.; number of sq. yd. of plastering = 4,416.5-5-9 = 490.72. 
Ans. 

(165) A triangle with three equal angles has three 
equal sides, and is, therefore, an equilateral triangle. 

(166) A triangle with two equal angles has two equal 
sides, and is, therefore, an isosceles triangle. 

(167) No, since the sum of the two shorter sides is not 
greater than the third side. 

(168) Draw a line B D from the vertex perpen¬ 
dicular to the base. Fig. 2. It will divide 

the base into two equal parts, as shown. In 
the right-angled triangle A B the hypot¬ 
enuse A B = 6y and side A D = 3; hence, 
by rule 50, Art. 385, B Dy the altitude = 

- 3’ = >/^ = 5.196 ft. Ans. 

(b) Area = ^ ^ = 15.588 sq. ft. Ans. 

(169) The sum of the three angles in any triangle = 2 
right angles, or 180°. In the given triangle, the sum of two 
angles = 23° + 32° 32' = 55° 32', and the third angle = 
180° - 55° 32', or 

180° = 179° 60' 
subtracting, 55° 32' 

124° 28'. Ans. 

(170) In Fig. 3 we have the pro¬ 
portion AD : D E \\ A B : B CyVn 
which A D = 10 in., A B = 24: in.y and 
.5 (7 = 13^ in., to find D E. 

. Substituting the given values, 

10 : DE :: 24 : 13^, or 




DE = 


10 X 13.5 
24 


= 5.625 in. Ans. 


( 171 ) A line drawn diagonally from one corner to the 
opposite one would form the hypotenuse of a right-angled 










98 


MENSURATION. 


triangle, whose two sides are 39 and 52 f eet. By rule 49, 
Art. 385, the length of the diagonal = 4 /52^ + 39^* = 65 ft. 

Ans. 


(172) Using rule 50, Art. 385, the required dis¬ 
tance = 4 / 24 " - 8" = 4/576 - 64 = 22.627 ft. = 22 ft. 7i in. +. 

Ans. 


(173) 

(174) 


See Art. 386. 


{a) Using rule 52, Art. 386, length of base = 
2 X 200 


20 


: 20 in. Ans. 


{b) A perpendicular let fall from the vertex to the base 
will divide the triangle into two equal right-angled triangles, 
of which the perpendicular side is 20 inches and the hori¬ 
zontal side 10 inches. Hence, according to rule 49, Art. 
385, the hypotenuse = 4 / 20 "* -j-10^ = 22.36 in. = the length 
of one of the equal sides of the given triangle. Ans. 


(175) An equilateral heptagon has seven equal sides; 
hence, the sum of all the sides, or the perimeter, = 7 X 3 = 21 in. 
Ans. 


(176) A regular decagon has 10 equal sides; hence, the 
length of one of the sides = 40 -4-10 = 4 inches. Ans. 


(177) 

389, we have interior angle 


A dodecagon has 12 sides, and by rule 53, Art. 
180 X (12 - 2) _ 


la 


(178) Divide the pentagon into five equal isosceles tri¬ 
angles, as shown in Fig. 4, by drawing 
a line from the center to each angle. 
The area of one of the triangles, as 
ABC, = 43 -4- 5 = 8.6 sq. in. We 
have given, therefore, the area and 
base B C oi the triangle A B C,to find 
its altitude A D. Using rule 52, 

Art. 386, A D = = 3.44 in., 

5 

the perpendicular distance from the center to one side. Ans. 













MENSURATION. 


99 


(179) See example, Art. 389. The process is simply to 
find one of the angles of the polygon, and then to divide it by 
2. By rule 53, Art. 389, one of the interior angles 

= !22^4tl)=i35-. 

o 

This divided by 2 = G7|-°. 

Ans. 

(180) Divide the 
figure into two triangles 
and one trapezoid, as 
shown in Fig. 5. The 
dimensions of the differ¬ 
ent parts can then be 
found by measurement. 

In obtaining the areas, fig. 5. 

it will be easier to use decimals than common fractions. 



Area of triangle A 

Area of triangle B = = . 188 sq. in. 

2 Z 

Area of trapezoid = X J ^ - X .75 = 1.25 X 

.75 = .938 sq. in. 

Hence, the area of the whole figure = . 656 + • 188 + • 938 = ' 
1.78-f-sq. in. Ans. 


(181) An angle inscribed in a circle is measured by 
one-half the intercepted arc. In this case, the angle 
intercepts one-fourth the circumference, and is measured by 


one-eighth the circumference, or by 360° x ^ = 45^ 

o 


Hence. 



there are 45° in the angle. Ans. 

(182) Since this is a regular 
hexagon, it may be inscribed in a 
circle (Fig. 6), and the radius of 
the inscribing circle will be equal to 
one side of the hexagon. Since the 
diameter E F = Z inches, the radii 


Lo4C. 




















100 


MENSURATION. 


A B and A C, and the side B C each = 1 inch, and the tri 
angle A B C vs, equilateral. Draw the line A D perpen¬ 
dicular to the side AU; it will bisect B C. Then, in the 
right-angled triangle A D B, A B = 1^ and B D — j-", to fi nd 
A D, According to rule 50, Art. 385, A D = i/V — .5^ = 
= .860''. Hence, the distance between two opposite 
sides of the hexagon = A D X ^ = .866 X 2 = 1.732". Ans. 

(183) In Fig. 7, we have the pro¬ 
portion A/: /A, in which 

B/=G, and ff / = I ot I/IC='^=Q. 

Substituting, 6:9::9:/y4, ov IA — 
81 

— =13.5in. Hence, the diameter^ 

6 

fig. 7. +.5/= 13.5-f 6 = 19.5 in. Ans. 

(184) If the diameter AB — ZZ\ ft. and 7^=8 ft., 
A 1= 32-J- — 8 = 24^ ft. Then, from the proportion of the 
last example, 8 : HI:: HI: 24.5, whence 777= |/8 X 24.5 = 
4/196 = 14 ft., and 7/Ar = 2 X 14 = 28 ft. Ans. (See 
Art. 400.) 

(185) Diameter = circumference 3.1416. By rule 
56, Art. 403, diameter of tree = 7.8543.1416 = 2|- ft. 
Ans. 

(186) One mile = 5,280 feet. The circumference of 

72 V 3 1416 

the wheel in feet =--- = 18.8496. (See rule 55, 

12 

Art. 402.) Number of revolutions in going one mile = 
5,280 -^- 18.8496 = 280.112. Ans. 

(187) Using rule 58, Art. 405, area = diameter 
squared X.7854. 6.06’= 36.7236; 36.7236 X.7854 = 28.8427 
sq. in. Ans. 

(188) By rule 59, Art. 406, inside diameter = 

= 12", Since the pipe is thick, the outside 
diameter must be f" x 2 = 1^" more, or 12 -f IJ = 13^". Ans 











MENSURATION. 


101 


(189) Since the radius of the circle = 6 in., its diam¬ 
eter = 12 in., and its circumference = 12 X 3.1416 = 37.6992 
in. There are 360° in the circumference, and the length of 

• 1 o 

an arc of 12° = 37.6992 X = 1-25664 in. Ans. 

360 

(190) The area of a circle 22 inches in diameter = 22* 
X .7854 = 380.1336 sq. in. (See rule 58, Art. 405.) Area 
of a circle 21 inches in diameter = 21* X .7854 = 346.3614 
sq. in. Hence, the area of a flat ring whose outside 
diameter = 22 in. and inside diameter = 21 in. is 380.1336 — 
346.3614 = 33.7722 sq. in. Ans. 


( 191 ) In the formula of rule 61, Art. 408 , 


"7-.608, h=- the height of the segment = 5 in., and 


3 

D =the diameter of the circle=56 in. Hence, the area of the 
segment 


4X5 




100 


^|/10.592 = 33iX 3.255= 


108.5 sq. in. Ans. 


(192) In Fig. 8, let A U represent a section of the 

largest square bar that can be planed 
from the round bar. A B and C D 
each = 2 in., and in the right-angled 
triangle A O the sides A O and C O 
each = 1 in. The hypotenuse A C 
|/IUjPU==|/^= 1.4142 in. Ans. Hence, 
the largest square bar that can be 
planed from the round bar is 1.4142^ 
square. fig. a. 

(193) The area of a circle 15 in. in diameter = 15* X 
.7854= 176.715 sq. in. Hence, the area of a sector of this 

• 1 u 1 • -iw-iio -inn ry-i 12^ 2,208.937 

circle, whose angle is 12^ , = 176.715 X -^ = —ggQ— = 

6.1359 sq. in. Ans. (See rule 60, Art. 407.) 

(194) {a) The side of a square whose area = 103.8691 

sq. in. = 103.8691 = 10.1916 in. Ans. 













102 


MENSURATION. 


(3) By rule 59, Art.. 406, the diameter of a circle 

having the same area = y ■—— = 111 Ans. 

(c) Perimeter of the square = 10.1916 X 4 = 40.7664 in.; 
circumference of the circle = 11.5 X 3.1416 = 36.1284 in.; 
difference = 40.7664 — 36.1284 = 4.638 in. Ans. 

(195) With 5 in. allowed for lap, the length of the 
plate which forms the shell is 46 — 5 = 41 in. This is the 
circumference of the shell. By rule 56, Art. 403, the 

41 

diameter corresponding to this circumference is - = 

13.05 in. Ans. 

(196) The convex area is the area of the outside sur¬ 

face, not including the area of the ends. The circumference 
of the base = 26 X 3.1416 = 81.6816 in. This reduced to 
feet, since the area is to be in feet, = 81.6816 12 = 6.8068. 

Using rule 62, Art. 416, convex area = 6.8068 X10^ = 
71.4714 sq. ft. Ans. 

(197) The perimeter of the base = 4 X 6 = 24 in. = 2 
ft. Convex area = 2 X 12 = 24 sq. ft. The area of the 

bases is found as follows: In Fig. 9, 
A B = 4: in. and A C = 2 in.; since this is 
a regular hexagon, A O = A B = 4 in. 
By rule 50, Art. 385, O C = |/4" — 2* = 
4/I2 = 3.4641 in.; area of triangle 
A O B ^ ^ X 1-^641 ^ _ 

area of base = 6.9282 X 6 = 41.5692, and 
the area of both bases = 41.5692 X 2 = 83.1384 sq. in. This 

00 1QQ4. 

reduced to square feet = —— = .5774. Hence, the area 

of the entire surface of the column is 24 + .5774 = 24.5774 
sq. ft. Ans. 

(198) The cubical contents in cubic inches = area of 
base in square inches X altitude in inches. The area of the 
base in the last example was found to be 41.5692 sq. in.; 











MENSURATION. 


103 


altitude = 12 X 12=144 in. Hence, the cubical contents= 
41.5692 X 144 = 5,985.9648 cu. in. Ans. 


(199) This example is solved by combining the rules 
for the circular ring (see example. Art. 406) and for the 
cylinder. To obtain the area of one end of the tube, we have 
4“ X .7854 = 12.5664 = area of a circle 4 inches in diameter; 
3.73^ X .7854 = 10.9272 = area of a circle 3.73 inches in 
diameter; difference = 12.5664— 10.9272 = 1.6392 = area of 
one end of the tube. The cubical contents = 1.6392 X 12 
= 19.6704 cu. in.; the weight = 19.6704 X . 28 = 5.5, or 5^- lb. 

Ans. 

( 200 ) This example is done exactly like the one in Art. 
417, and the solution is given here without explanation. 

(a) In the formula 'of rule 61, Art. 408, 


i^i/£ 

r *T* 


— .608, /i in this case = 18, and D = 60. 


Substituting, area = 

.608 - ^ ^ 4/3.333-.G08 = 432X-/2.725= 

3 18 o 

432 x 1.65 = 712.8 sq. in. This reduced to square feet 
= 712.8 -f- 144 = 4.95. Hence, the steam space = 4.95 X 
16 = 79.2 cu. ft. Ans. 


(d) Total area of one end of boiler in square inches = 
60* X .7854 = 2,827.44. From this is to be subtracted the 
area of the tube ends and of the segment found above. 

Area of ends of tubes = 3.5* X .7854 x 64 = 615.75 sq. in. 
Area of segment = 712.8 sq. in. 


1,328.55 sq. in. 

Area of water space = 2,827.44 — 1,328.55 = 1,498.89 
sq. in. 

Contents of water space = 1,498.89 X 16 X 12= 287,786.88 
cu. in., and 287,786.88 231 = 1,245.83, number of gallons, 

or say 1,246 gal. Ans. 


(201) (a) Area of piston = 19* X .7854 = 283.529 sq. in., 
or 1.9689 square feet (rule 58, Art. 405). 









104 


MENSURATION. 


Length of stroke plus the clearance = 1.14 X 2 ft. (24 in. 
= 2 ft.) = 2.28 ft. 

1.9689 X 2.28 = 4.489 cubic feet, or the volume of steam 
in the small cylinder (rule 63, Art. 417). 

(d) Area of piston = 31* X .7854 = 754.7694 sq. in., or 
5.2414 square feet. 

Length of stroke plus the clearance = 1.08 X 2 = 2.16 ft. 
5.2414 X 2.16 = 11.321 cubic feet, or the volume of steam 
in the large cylinder. Ans. 

11 

(^r) Ratio = - or 2.522 : 1. Ans. 

’ 4.489 

(202) Let the equilateral triangle ABC, Fig. 10, rep¬ 
resent the base of the pyramid. By rule 
50, Art. 385, the altitude A D of the tri¬ 
angle = 4/10*^^^= 4/^ = 8.6602 in., and 
according to rule 51, Art. 386, the area 

^ of the triangle =- - -= 43.301 sq. in. 

Fig. 10. Using rule 65, Art. 423, volume of pyra- 
1 4 Q QA 1 V 10 

mid = area of base X ^ altitude = —-- =144.336 cu. 

in. Ans. 

(203) In Fig. 11, let O H he the altitude and 0£ the 
slant height of the pyramid. 

Connect points H and A, 
forming the right-angled tri¬ 
angle O H E, in which we 
have to find OH. Since it 
is a right pyramid, point H 
will fall at the center of the 
base A B C D, and, hence, the 

line HE = ^A B, or 8 in. ; 

OE =z 26 in.; by rule 50, 

Art. 385,<9A^=4/25*-8*= ^ 

4^561 = 23.6854 in. Ans. 




Fig. 11. 












MENSURATION. 


105 


( 204 ) The area of the convex surface = circumference 

of base x ^ slant height = 18.8496 X — = 94.248 sq. in. 

(See rule 64 , Art. 422.) The area of the entire sur¬ 
face = 94.248 sq. in. + the area of the base. The diameter 

of the base = ■ = 6 in.; hence, the area of the base = 

6’X .7854 = 28.2744 (rules 56 and 58 , Arts. 403 and 
405 ); therefore, the area of the entire surface = 94.248 + 
28.2744= 122.5224 sq. in. Ans. 

( 205 ) Using rule 65,Art. 423,volume=area of base x ~ 

O 

altitude = 28.2744 X |- = 84.8232 cu. in. Ans. 

O 

. ( 206 ) The vat has the form of an inverted frustum 
of a pyramid. Area of larger base = 15^ = 225 sq. ft.; 
area of smaller base = 12^ = 144 sq. ft. Hence, by rule 
67 , Art. 427 , the contents of the vat in cubic feet = 


(225 + 144 + |/225X144) 


y = (369 + 180)X^ 


549 X~ = 


2,013 cu. ft. This should be reduced to cubic inches by 
multiplying by 1,728, the number of cubic inches in a 
cubic foot. 2,013 X 1,728 = 3,478,464 cu. in. Since there 
are 231 cubic inches in a gallon, the number of gallons 


^ 4.7« Afii. 

that the vat will hold = ■ ’ ■ ' -= 15,058.29. 

231 


Ans. 


( 207 ) The pail is in the form of a frustum of a cone. 
Area of larger base = 12“ X .7854= 113.0976 sq. in. Area 
of smaller base = 63.6174 sq. in. Hence, the contents in 
cubic inches = 

(113.0976 + 63.6174+4/113.0976 X 63.6174) X ^ = 
(176.715 + i/7,194.9753)^ = (176.715 + 84.8232) X^ = 


261.5382 X 


11 


958.9734. 


The contents of the vat in cubic inches were found in the 


H.a. /V.—7 







106 


MENSURATION. 


last example to be 3,478,464. Hence, the number of pails of 
water required to fill the vat = 3,478,464958.9734 = 
3,627.28. Ans. 

( 208 ) By rule 66 , Art. 426 , the area of the convex 
surface = half the sum of the perimeters of the upper and 

' . 48 36 

lower bases X the slant height, or- - -X 32 = 42 X 32 = 

/i 

1,344 sq. in. 

( 209 ) See note following the question. 

170.5 

Outside diameter of upper base = = about 54.27 

in.; inside diameter = 54.27 — 2.5 = 51.77 in.; area of upper 
base = 51.77“ X .7854 = 2,105 sq. in., nearly. 

190 

Outside diameter of lower base = ^ 1410 ~ “ 30.48 in., 

nearly; inside diameter = 60.48 — 2.5 = 57.98 in.; area of 
lower base = 57.98“ X .7854 = 2,640 sq. in., nearly. 

Apply rule 67 , Art. 427 . Contents of the tank in cubic 
_ 7 V 12 

inches = (2,105 + 2,640 + l/2,105 X 2,640) X —- = (4,745 

o 

+ 2,357) X 28 = 7,102 X 28 = 198,856 cu. in. Hence, the 
number of gallons = 198,856 -J- 231 = 860.8. Therefore, in 
round numbers, the tank will hold 861 gallons. 

( 210 ) (a) By rule 68 , Art. 429 , area of the surface = 
22.5“ X 3.1416 = 506.25 X 3.1416 = 1,590.435 sq. 'in. Ans. 

(5) Using rule 69 , Art. 430 , the cubical contents = the 
cube of the diameter X .5236 = 11,390.625 X .5236 = 
5,964.1313 cu. in. Ans. 

( 211 ) Having given the area of the surface, to find the 
volume we must first obtain the diameter. The process is 
just the reverse of finding the surface when the diameter is 
given. Hence, the diameter = 4/201.0624-7-3.1416 = 4 /^ — 
8 in. By rule 69 , Art. 430 , volume = 8 ® X. 5236 = 268.0832 
cu. in. Ans. 








MENSURATION. 


107 


( 212 ) {a) Given O B =:~^ or S inches, and O A = 

or 6 |- inches, to find the volume, area, and weight (see 
Fig. 12): 

Radius of center circle equals — ^ or 7^ inches. 

Z 

Length of center line = 2 X 3.1416 X 
7^ = 45.5532 inches. 

The radius of the inner circle is 6 ^ 
inches, and of the outer circle 8 inches; 
therefore, the diameter of the cross- 
section on the line A B is 1|- inches. 

Then, according to rule 70 , Art. 431 , 
the area of the ring is 1 ^ x 3.1416 X 45.553 = 214.665 square 
inches. Ans. 

Diameter of cross-section of ring = 1 -^ inches. 

Area of cross-section of ring = (l-^-)’ X .7854 = 1.76715 sq. 
in. Ans. 

By rule 71 , Art. 432 , volume of ring = 1.76715 X 45.553 
= 80.499 cu. in. Ans. 

(d) Weight of ring = 80.499 X .261 = 21 lb. Ans. 



( 213 ) The volume of the ring equals the product of its 
cross-sectional area and its length on line D; therefore, 
144.349 

cross-sectional area = - -- r — = 7.069 sq. in. 

20.42 

/7.069 

The diameter A B = y = 3 in. 

.7854 

Area of surface = circumference of circle, of which A B 
is diameter, X length = 3.1416 X 3 X 20.42 = 192.45 square 
inches. Ans. 







1 


1 

,'i 

% 



t 


ELEMENTARY ALGEBRA 

AND 

TRIGONOMETRIC FUNCTIONS. 

(QUESTIONS 314-368.) 

(214) See Art. 531. 

(215) (a) 3.r -j- 6 — = 7^. Transposing 6 to the 

second member, and 7^ to the first member (Art. 575), 

3x-2jt:-7x= -6. 

Combining like terms, — Qx = — 6; 
whence, x = 1. Ans. 

{b) bx - {3x -7) = ^x- (ex - 35). 

Removing the parentheses (Art. 482), 

5;ir - 3;r + 7 = 44r - 6;r + 35. 

Transposing 7 to the second member, and 4;r and — ex to 
the first member, 

5x — 3x — 4:X-}- ex = 35 — 7. (Art. 575.) 

Combining like terms, 4:X= 28; 
whence, = 28 -f- 4 = 7. Ans. 

(c) (x -|- 5)“ — (4 — xy = 21x. 

Performing the operations indicated, the equation becomes 

.r* + 10;r -f 25 — WSx — x^ = 21.r. 
Transposing, .r" x"" + 10;r %x — 21x = IG — 25. 

Combining like terms, — Zx = — 9. 

Dividing by — 3, x =3. 

For notice of copyright, see page immediately following the title page. 



110 


ELEMENTARY ALGEBRA AND 


(216) 

^2 _ -1 )2«6 _ 4a= - 5a^ + Sa'^ + lOa^ 4- 7^^ + 2(2^3 _ 2a'^ - 3a - 2. 

2«6 - 2a^ - 4a^ - 2a^ Ans. 

— 2a^ — a^ + 5<a!3 + lOa^ 

— 2«5 + 2a^ + 4^3 2«2 

— Srt** + «3 ^ 3^2 _|_ 7^ 

- 3^4 +3^3 6«2 + 3« 

-2«3 4- 2«2_,_4^_^2 
— 2^3 4- 2a'^ 4- 4^^ 4- 3 


(217) (a) 2 -f 4^ - 5a^ - 6a^ 

7a^ 

Ua^ + 28^?" - 35a^ - 42a\ Ans. (Art. 493.) 

(^) 44 ;^ _ 4y + 6 ^" 

34r> 

124r> - 12.ry + 18;r>^^ Ans. 

(c) 8^-^ 5c —2d 

6a 

ISad -I- 80ac — V2ad. Ans. 


(218) 

Then, 


Let 4r = number of miles he traveled per hour. 

48 . . , , . 

— = time it took him. 

X 


48 

X-j- 4: 


= time it would take him if he traveled 
4 miles more per hour. 


In the latter case the time would have been 6 hours ; 
hence the equation 


48 ^ 

ji: ^ 4 

Clearing of fractions, 48 = 6 {x 4). 
Dividing by 6, 8 = ;r 4. 

Transposing, 8 — 4 = 4r. 

X = 4. Ans. 


(219) The square root of the fraction a plus b plus c over 
n, plus the square root of plus the fraction b plus c over 
plus the square root of the quantity a plus b, plus the frac- 










TRIGONOMETRIC FUNCTIONS. 


Ill 


tion c over plus the parenthesis a plus b, times r, plus a 
plus be. 

(220) {a) Writing the work as follows, and canceling 
common factors in both numerator and denominator (Arts. 
545 and 540), we have 

hp'^q 24;ry _ 

9 X 5 X 24 X X //"* X X ^ X X jk" _ ^mnxy 

8 X 2 X 90 X ;;/ X ;/ X X X ^ X ^ * 

(b) This problem may be written as follows, according to 
Art. 529, 

Sax 4- 4 _ ^ _ 

1 a(Sax -|- 4) {Sax + 4)* 

Canceling a and {Sax + 4), we have -— ^ -. Ans. 

(Sax —j— 4* 

( 221 ) See Arts. 602 and 603 . 

Sin 17° 28'= .30015. 

Sin 17° 27'= .29987. 

30015 — 29987 = 28, difference for 1'. 

X -11- = 17, difference for 37". 

29987 + 17 = 30004. 

Hence, locating the decimal point, 

Sin 17° 27' 37" = .30004. 

Cos 17° 27'= .95398. 

Cos 17° 28' = .95389. 

95398 — 95389 = 9, difference for 1'. 

9 X 14 = 6, difference for 37". 

95398 — 6 = 95392. 

Hence, Cos 17° 27' 37" = .95392. 

Tan 17° 28' = .31466. 

Tan 17° 27'= .31434. 

31466 — 31434 = 32, difference for 1' 

32 X 14 = 20, difference for 37". 

31434 + 20 = 31454. 

Hence, Tan 17° 27' ST = .31454. 









lia ELEMENTARY ALGEBRA AND 

Sin 17° 37'37" = .30004. ^ 

Cos 17° 37' 37" = .95393. V Ans. 

Tan 17° 27' 37 ' = .31454. ) 

(222) Let ;i' = the capacity. 

Then, x — 4:2 = amount held at first. 

7(x - 4:2) = X. 

7x - 294 = ;ir. 

Qx = 294. 

= 49 gallons. Ans. 

(223) ~l~ <^) + L be -j- cd ^ 

{a-\-b)c ac^bc ^ ’ 

. 1 1 ^ “L ^ ^-11 ^ 

IS common to each term, we have-,— 7 — = 1 H-j—»• 

’ a-\- b a b 

Ans. 

(224) The solution is exactly similar to that of example 
221 , preceding. 

Sin of 63° 4'51.8"= .89165. 

Cos of 63° 4'51.8" = .45274. 

Tan of 63° 4' 51.8" = 1.96949. 

(225) (<;-*)-» = £* Ans. 

{m = -4-5. Ans. 

m^ir 

{cd~'^Y = or or Ans. (Art. 565.) 


( 226 ) --- ^7—1— 7 H- 7 -—denominator 

2 — X 2 ^ ^ — 1 

of the third fraction were written 4 — x\ instead of x^ — 4, 
the common denominator would then be 4 — 


By Art. 531 , 
Hence, 


16.r 


x^ - 4 : 

3 + 2-r 2 — 3x 


x\ IQx-x- 

becomes — 


16^: 


;r-' + 4 


16;r 


2 -T 


, when reduced to a 


common denominator, becomes 














TRIGONOMETRIC FUNCTIONS. 


113 


(3 + 2jtr) (2 + ,r) - (2 - 3.r) (2 -.r) - (16.r - _ 

4-~ 

(6 + 7;ir + 2.r^) - (4 - 8.r + 3^:*) - (lG;r - 

Removing the parentheses (Art. 482), we have 

6 -f 7;ir 4- 2;r’ - 4 + Sx - 3;r" - Ux-^x^ 

4- x^ 

Combining like terms in the numerator, we have 
2-.r 
4 - ;r"‘ 

Factoring the denominator (Art. 523), we have 
2-.a: 

i^ + x) ( 2 -xy 

Canceling the common factor (3 — x), the result is equal to 

ac — b" — 2ac 

- r’ - tiad ^ a’c + aV + ae' ' 

Arranging the terms, we have 
ac -\r 

a' - -lab -Yb'‘-c‘ ^ rtV+aV + ar* ’ 
which, by the use of parentheses, becomes 

a’-q-rtr + r’ (a’- + f’) - 

(a^ - 'iab + b') -c'‘^ + a‘c‘ + ac' ' 

By Art. 518, we know that — 2ab + and — 2ac + 
are perfect squares, and may be written {a — bf and (a — cy. 

Factoring + ab by Case I, Art. 516, we have 

-j- ac b {a — cy — b _ 

(^— by — b ^ acia" acb) ~~ 

a^-ytic-{-b _ {a — c — b){a — c-^b) 

(a — b — c) {a — b-fTj ^ ac{a^ + + ^') 

(Art. 523.) 



















114 


ELEMENTARY ALGEBRA AND 


Canceling common factors and multiplying, we have 

a — c A- b aA-b — c . 

7-—?-Trn—7* 

{^a — b c)ac acya — b c) 

(228) If none of the terms is similar, the subtraction 
of one expression from another may be represented only, by 
connecting the subtrahend with the minuend by means of 
the sign —. Thus, if it is required to subtract ha^b — la^b’' -|- 

from o' — b\ the result will be represented by a* — b* 
— {oaAb — 7a^b^ + 5ab^), which, on removing the parenthesis 
(Art. 482), becomes o' — b^ — hoAb + loAb"^ — 6ab^. From 

this result subtract Za* — ^loAb -f- ^oAb"^ -R ~ 

inmuend. 

- + ^b"^ + 4:a^b — 6a^b'^ — 5ab^ subtrahendy with signs changed. 

~ 2 ^“^ + 2 ^“* — d^bdb^ — lOab^ remainder. (Art. 4T9.) 

Or, — — oAb -f- oAb"^ — l^ab^ + Ans. arranged 

according to the decreasing powers of a. 

(229) .27038 = sin 15° 41' 12.9". (Art. 606.) 

.27038 == cos 74° 18' 47.1". 

2.27038 = tan 66° 13' 43.2". 

(230) {d) If the work done on a piston by the confined 
gases be considered positive, then the work done by the 
piston in pushing the gas out of the exhaust-port may be 
considered negative. 

(b) While in arithmetic we can add and subtract only 
positive quantities, in algebra we can perform these opera¬ 
tions on both positive and negative quantities. 

(231) {a) The value of <3:° is 1. (Art. 503.) 

(^) —1 = Ans. (Art. 565.) 


( 232 ) 


(a — b) _ (a — b)~-c 
c-\-{a-y-b)~ c~{-{ay- b)' 


(Art. 531 .) 


( 233 ) {a) By Art. 530 , the reciprocal of ff = 1 = 

1 X |-|- = If. Ans. 







TRIGONOMETRIC FUNCTIONS. 


115 


{b) Since, by Art. 530, a number may be found from 
its reciprocal by dividing 1 by the reciprocal, the number = 
1 700= .0014f. Ans. 

(-^3-4:) (^) 3^ — %b -f~ 3^ 3^ — 2^ -f“ 

2^ — c becomes — 2^: + 8/^ + c 

a Qb 

when the signs of the subtrahend are changed. Now, adding 
each term (with its sign changed) in the subtrahend to its 
corresponding term in the minuend, we have (— 2^) + (3«) = 
^; ( + %b) + ( — 2^) = -|- 6^; ( -J- ^r) + (3c) = + 4c. Hence, 
^ + 6^ + 4c equals the difference. Ans. 

{b) 

— 3;r-b' + + 2.ry 

|3eQoi^es 

x^ — 3xy + 2xj^^ —j'^ + xy‘‘ 

when the signs of the subtrahend are changed. Adding 
each term in the subtrahend (with its sign changed) to its 
corresponding term in the minuend, we have — 3xy + 
2;rK® — y which, arranged according to the decreasing 

powers of equals x^ — 3xy + xy^ 

(c) 14^^ + 4<^ - 6c - 3^ 

11^ — 2<^ + 4c — 4^ 

On changing the sign of each term in the subtrahend, 
the problem becomes 

14^ + 4^ - 6c — 3^ 

— 11a-{-2b— 4c-|-4^ 

3a-{-6b- 10c + d 

Adding each term of the subtrahend (with the sign 
changed) to its corresponding term in the minuend, the 
difference, or result, is 3<3: + 6^ — 10c -|- d. Ans. 

(235) The numerical values of the following, when 
^ = 16, ^ = 10, and .r = 5, are: 








116 


ELEMENTARY ALGEBRA AND 


(a) {ab'^x + %abx) 4^ = (16 X 10^ X 5 + 2 X 16 X 10 X 5) X 
4 X 16. It must be remembered that when no sign is 
expressed between symbols or quantities, the sign of multi¬ 
plication is understood. 

(16 X 100 X 5 + 2 X 16 X 10 X 5) X 64 = (8,000 + 1,600) X 
64 rr 9,600 X 64 = 614,400. Ans. ' . 


{P) 



b — X 
X 


16 


2l/^- 
10^+1 = 


2 X 10 X 5 
16-10 
96 — 100 + 6 
6 



1 

X' 


Ans. 


(<;) (I, _ 4/a) (;r*_ b’‘) {a‘-b'‘) = (10 - vTe) X (5’-10’) X 

6’ - 10’) = (10 - 4) (125 - 100) (25C - 100) = 6 X 25 X 

The sum of the coefficients 
of the positive terms we find 
to be + 13, since ( + 3) + 
(+6) + (+4) = (+13). 

When no sign is given be¬ 
fore a quantity, the + sign 
must always be understood. 
The sum of the coefficients of 
the negative terms we find to be —17, since (— 9) + (— 5) + 
(— 3) = (— 17). Subtracting the lesser sum from the 
greater and prefixing the sign of the greater sum ( — ) 
(Art. 471, Rule II), we have (-[- 13 ) + ( — 17 ) = — 4. 
Since the terms are all alike, we K&ve only to annex the 
common symbols xyz to — 4, thereby obtaining — \xyz for 
the result, or sum. 

(b) + ‘^ab + 4(^^ 

+ b"^ 

— + hab — b"^ 

18^" - ^Oab - 19^ 
l^a^ — ‘dab + 20(^'* 

39^** — ^^ab + 5^k Ans. 


When adding polynomi¬ 
als, always place like 
terms under each other. 
(Art. 478.) 

The coefficient of a^ in 
the result will be 39, since 
(+14)+ (+18)+ (-!) + 


156 = 23,400. Ans. 

(236) {a) ^xyz 

— dxyz 

— bxyz 
Qxyz 

- ^xyz 
dxyz 

— ^xyz Ans. 








TRIGONOxMETRIC FUNCTIONS. 


117 


(+ 5) -j- (+ 3) = 39. When the coefficient of a term* is not 
written, 1 is always understood to be its coefficient. 
(Avt. 442.) The coefficient of al? will be — 24, since 
(~ 3) + (- 20) + (+ 5) + (- 8) 4- (+ 2) = - 24. The 
coefficient of will be (-f- 20) -|- (— 19) + (—!) + (+ 1) + 
(+ 4) = -j- 5. Hence, the result, or sum, is — 24:al? + 5^^ 

(c) ^inn 4- ^ab — 4r 

4" — \ab ‘dm^ — 4/ 

Qmn — ab — — 4/. Ans. 

(237) {a) See Art. 445. 

{b) In multiplication, coefficients are multiplied, and 
»jxponents are added. In division, the coefficients of the 
dividend are divided by those of the divisor, and the expo¬ 
nents of the divisor are subtracted from those of the divi¬ 
dend. See rules of multiplication and division. 

(c) See Art. 487. 

(238) The side />’ C = - XC’, or B C = 

/17.69’ — 9.75“ = 4/^17.8736 = 14 ft. 9 in. To find the 
angle BAG, we have 

cos BA C — 44 or cosB A C — = .55116. (Art. 595.) 

AB' 17.69 ' ’ 


.55116 = cos 56° 33' 13.5'. 

Angle A BC= 90° - angle BA C, or 90° - 56° 33' 12.5’ = 


33° 20' 47.5'. 

Side C = 14 ft. 9 in. 


( 239 ) 


Angle B AC 33' 12.5'. V Ans. 
Angle ^ .5 C = 33° 26' 47.5’. ) 

, 9r+20 4(;i-- 3) , J." 

-^- = -^ 7 ^ + 4 - 


When the denominators contain both simple and com¬ 
pound expressions, it is best to remove the simple expres¬ 
sions first, and then remove each compound expression in 
order. Then, after each multiplication, the result should 
be reduced to the simplest form. 








118 


ELEMENTARY ALGEBRA AND 


Multiplying both sides by 36, 

144 (.r - 3) 


9;i: + 20 = 


or 


54: — 4 

1444- - 432 
54- — 4 


+ 


20 . 


Clearing of fractions, 

1444- - 432 = 1004- —80. 
Transposing and combining, 

444'= 352; 

whence, 4' = 8. Ans. 


(/') 


ajf - 


_ 1 

-= — becomes, when cleared cf fractions, 


2aAr — 3a -j- If jr = 1. 
Transposing and uniting terms, 

2ajir -j- = 3a -j- 1. 

Factoring, (2a = 3a -\-l; 

3a 1 

whence, r = -r —,— r- 

’ 2a-{- d 


Ans. 


(c) am — b -r-d-=0 becomes, when cleared of 

^ ’ h ni 

fractions, 

abiff — b'^m — amx bx = 

Transposing, bx — amx — b'^m — abnt^. 

Factoring, (b — am)x = bm(b — am ); 

, bm{b — am) , . 

whence, r = —rr -= bm. Ans. 

(b — am) 


(240) (a) a square 4- square, plus two a cube b fifth, 

minus the parenthesis a plus b. 

(b) The cube root of 4 :, plus jr times the two-thirds power 
of the parenthesis a minus n square. 

(c) The parenthesis m plus ;z, times the square of the 
parenthesis 7n minus times the parenthesis minus the 
fraction 71 over two. 







TRIGONOMETRIC FUNCTIONS. 


119 


(241) {a) a*iad; 4:a''— da''7ax. 

(d) Since the terms are not alike, we can only indicate the 

sum, connecting the terms by their proper signs. (Art. 47t>.) 
(r) Multiplication: means4(Art. 441.) 

(242) (^) 45.ry" - 00,ry - 3C)0.t'y = 

45.iry(.r'y — — 8j). Ans. (Art. 516.) 

(I?) %abcd -j- = {ab + cd^- Ans. (Art. 519.) 

(c) (a -j- by — (c — dy = (a b -j- c —d) {a b — c d). 
Ans. (Art. 523.) 

(243) {a) On removing the vinculum, we have 

2^ — \'^b [4^' — 4^? — yia -f- 2^)] + [Ih? — — r]}. 

(Art. 482.) 

Removing the parenthesis, 

‘ia — y^b [4r — 4^? — 2^ — 2/^] + [Wa — b —c^ J. 
Removing the brackets, 

'‘Za — \^b -\- — ia — Za — Zb da — b — c]. 

Removing the brace, 

Za — db — 46 ' + 46 ? -|- 26 ? + 2 /^ — ^ 

Combining like terms, the result is 5 a — dc. Ans. 

{b) Removing the parenthesis, we have 
76? — { 36? — [26? — 56? -f 4:6?] }. 

Removing the brackets, 

76? - {36? - 26?+ 56? - 46?). 

Removing the brace, 

76? — 36? + 26? — 56? + 46?. 

Combining terms, the result is 5 a. Ans. 

(c) Removing the parenthesis, we have 

6? — j 2<^ + [36: — 36? — 6? — ^] + [26? — b — /:]}, 
Removing the brackets, 

6 ? — {2<^ + 36- — 36? — 6? — 6^ + 26? — 6 ^ — 6:}. 
Removing the brace, ' 

6? — 26^ — 3<: + 36? + 6? + 6^ — 26? + ^ + c. 

Combining like terms, the result is 3?? — 26:. Ans. 


120 


ELEMENTARY ALGEBRA AND 


{24-t) The hypotenuse A D = A C + Li C , or 
A B = /IV.S* + 21.1T = t/fSOToi = 27.57, nearly. Ans. 

Tan .5 = 4$= ^ = .82100. (Art. 596.) 
h L Zl.o 

.82100 = tan 39° 24' 23". 

Angle B — 39° 24' 23". Ans. 

Angle A = 90° - angle ^ 90° - 39° 24' 23" = 50° 35' 37". 

Ans. 

(2451 t - 

(^40) t- ■ 

In order to transform this formula so that may stand 
alone in the first member, we must first clear of fractions. 
Performing this operation, we have 

tW^s^ + tW^s^ ^ 

Transposing, we have 

- - tW^s^ - tlV,s^. 

Factoring, we have 

- = IVj,L, - {IV,s, + 


whence, 


{W,s,+ lV^)t-lV,s,L, ^ Ans. 


( 246 ) Let = the part of the work which they all can 
do in 1 day when working together. 

Then, since — = t + i > 

or, clearing of fractions and adding, 

15 = 30;r, and ^ 

Since they can do ^ the work in 1 day, they can do all the 
work in 2 days. Ans. 

( 247 ) {a) 1 1 i + ;r-l + .ar 




\ — X 

l + ^_ 

1 - 

1 1 

1 

1-l-x-l- 1 - X 

1 -x ' 

1 -j- X 

1 -x^ 

2 x 

1 - 


~l-x^ 

2 

= X. Ans. (Set 

















TRIGONOMETRIC FUNCTIONS. 


121 


a 


-\-b^ 

ab^ 


ab^ 


a a — b ifb — If (a — b) cfb — ah'" + b"" 

b^ ab ~ab^ Hb^ 


ab ab^ 

af^lf ^ a^b-ab’^^b^ b 


ab^ 


ab^ 


ah'" ab 

a -j- b 


b (of — ab-{- b'") 


b • 


Ans. 


(^) 




1 + 


ar+l 


■^-h 


3 — Jir 3;r + 3 


Ans. 


3 -Jtr (Art. 559.) 

(248) (a) 6a*b* + a^b^ - ^afb^ + 2 abc + 3. 

(b) 3 -1- 2abc + afb — '^afb 4- 

(^:) 1 ^ ax-\-of2af. Written like this, the a in the 
second term is understood as having 1 for an exponent; 
hence, if we represent the first term by af^ in value it will 
be equal to 1 , since of = 1 . Therefore, 1 should be written 
as the first term when arranged according to the increasing 
powers oi a. 

(249) {^) According to Art. 563, xl expressed radi¬ 
cally is 

3x^j ~^ expressed radically is 3 Vxj^; 


3 x^j/~h^ = 3 |/.rjr" V, since = xr*. Ans. 
Ar 
b^ 


(b) (See Art. 565.) a 'b^ + ~ ^ ^- 3 — 


a-\-b 


Ans. 


a (a -j- b) VI — 71 b' 

{c) \/x^ = x^. An^. \/x~* = x-^. Ans. 

{\/¥^y = (Rr^)’ = b^x^. Ans. 

2ax 4- 


(250) {a) 


;i' _ x{2a x) a — x 


(Art. 549.) 


B.a. ir.—s 





























122 


ELEMENTARY ALGEBRA AND 


Canceling common factors, the result is 


2a-\- 


a — x) a 

— a^x 


-{-axx'‘ 
x^ (a^ -j- ax -j- x^ 


^ a’^x — x^ 
a^x — ax’^ 


ax'‘ — 
ax^ — x^ 


(^) Inverting the divisor and factoring, we have 

^n{2in‘^n — 1) {2iffn -f-1) (2m^n — 1) 

(2m'‘n — 1) (2m^7t — 1) ^ 3n 

Canceling common factors, the result is 2ja^n + 1. 

(^) ^ + x^y (3 + simplified gives 

9x^ - 4y , 3x -^ 2j/ 

x^~f ‘ X—J7 • 

_4:1/* X y 

Inverting the divisor, we have ^ X -r^- 

x^ —X dx 2y 

celing common factors, the result is Ans. 

x-\-y 

(251) {a) 2x^ + 2;r* + 2x - 2 

jr -1_ 

2;r‘ + 2;r* + 2x^ — 2x 
- 2;r* - 2.r* - 2.ir + 2 
2x* _ 4;ir 2. Ans. 

{b) x"^ — ^ax -[- c 

2x —|— a 

2x^ — 8ax'‘ + 2^:.ir 

ax — 4^ X —1~ ac 

2x^ — '{ax^ + — 4^*.ir ac. ■ Ans. 

(c) - Za^b- 2b^ 

5<2* -j- ^<^b 

- 5a^ + Ua*b ~ lOa'^b^ 

- ^a"b _ + - 1 8ab 

^ ha^ + 6 ^ Y - 4 - 2^a^b’^ - {Sab 


Ans. 


Ans. 


Can^ 





















TRIGONOMETRIC FUNCTIONS. 133 

Arranging the terms according to the decreasing powers 
of a, we have 

- 5a' + Ga‘/? + 27a'l?' - lOa'i' - 18ad'. Ans. 


X X V 

(252) ia) -1-If the denominator of the 

' X — y y — X 

second fraction were written x —y, instead oi y — x^ then 

X — y would be the common denominator. 

By Art. 531, the signs of the denominator and the sign 

X _ y X _ y 

before the fraction -—^ may be changed, giving — —— 

We now have 


X X — y _x — X y _ y 

X — y X — y~ X — y ~~ x — y 


Ans. 


X* X X 

(b\ —^---I-—- — -. If we write the denomina- 

^ ’ x'^ — 1 ' x-j- 1 1 — X 

tor of the third fraction x — 1 instead of 1 — x, x^ — 1 will 
then be the common denominator. 

By Art. 531, the signs of the denominator and the sign 

before the fraction may be changed, thereby giving --. 

X X. 

We now have 


^ ^ = £l±£(£illl±^(£±i) 




3x^ 


-. Ans. 




X^-l 

3a — 4/^ 2a — ^ -j- ^ 13a — 4^ 


7 3 ' 12 

when reduced to a common denominator = 

12(3^ - 4^) - 28(2a - ^ + 7(13^ - 4c) 

84 

Expanding the terms and removing the parentheses, we 
have 

36^ — 48^ — 56a -f~ 28^ — 28^^ -f- Ola — 28^^ 


84 






















124 


ELEMENTARY ALGEBRA AND 


Combining like terms in the numerator, we have as the 
result, 


71 ^ - 20 /^ — 56c 
84 ' 


Ans. 


( 253 ) (a) Factoring each expression (Art. 519 ), we 

have 


+ 12;ry + 4/ = + 2/) + 2/) = (3;r* + 2y)^ 

Ans. 

(I?) 49 ^^ - 154 ^^^^ + = {7a^ - 11 ^^) (7a^ - lU^) = 

(7a^ - lUy. Ans. 

(c) 64A'y + 64;rr + 16 = 16 {2xf + l)^ Ans. 


(254) (a) Arrange the dividend according to the 

decreasing powers of x and divide. Thus, 

3;r - 1) 9;ir^ + 3;ir“ + .r - 1 ( 3x^ -f 2;r + 1. Ans. 

9^3 _ 3^2 

6x^-j- X 
6x^ - 2;r 

3;r- 1 
3:r- 1 

(fi) a-b) {o' ab — b"". Ans. 

— a'b 

a^b — ’^ab'" 
a^b — ab"^ 

- ab-^^b^ 

— ab"^ -f- 

(c) Arranging the terms of the dividend according to 
the decreasing powers of x^ we have 

7x-3)7x^- 24x^ + 58;r _ 21 (- 3;r + 7. Ans. 
7;r^- 3x^ 

— 21x^ -f 58;r 
-21;r=^+ 9x 


49;ir - 21 
49 jr — 21 












TRIGONOMETRIC FUNCTIONS. 


135 


( 255 ) See Arts. 435 and 436 . 

(256) (<r) 

1 I 3.r + + 4 

5x 5x 5x ' 


^ ' ;r -j- 4 


10 + 


41 


x-\-4. 


(Art. 555 .) 
Ans. (Art. 554 .) 


X + 4 ) 3x^ + ^x + 1 ( 3x — 10 + 
3x^ + 12x 


41 


X-{-4: 


-10x-{- 1 
- 10;r - 40 


41 

(257) The angle ^ = 90° — angle A = 90°— 65° 13' 29"= 
24° 46' 31". Ans. 

The side BC= AB x sin + = 5.5 yd. X sin 65° 13' 29" = 
5.5 yd.X .90796 = 4.9938 yd. (Art. 609.) 4.9938 yd.= 

14 ft. Ilf in. Ans. 

Side A C = A B X cos A = 5.5 yd. x .41906 = 2.3048 yd. 
= 6 ft. 11 in., nearly. Ans. 

(258) (a) {X* - 1) - (x^ + 1) = (x^ - 1) (x^ + 1)- 

(;r“ + 1) = — 1. Ans. (Art. 523.) 

(I?) — 2;ry+y = (.r" — j’')’. (Art. 519.) (.r"—y) = 

{x-y). (Art. 523.) 

Then (x^ — y^y = (x^ — y^) (x + y) (x — y). Dividing this 
latter quantity by (x — y) we have (x^ —+*) (^ + J^)- Ans. 

Note that x—y is a factor of (^’—j^*)^ and hence of 

- 2.iry +y. 

(259) (a) 

Reducing the last member to a simpler form, the equation 
becomes 

10;r + 3 Ox 


2 


= 10;ir - 10. 


Clearing of fractions by multiplying each term of both 
members by 6, the common denominator, and changing the 














126 


ELEMENTARY ALGEBRA AND 


sign of each term of the numerator of the second fraction, 
since it is preceded ,by the minus sign, we have 

20;r + 6 - ISx + 21 = 60;r - 60. 

Transposing terms, 20 x — ISx — 60jtr = — 60 — 21 — 6. 

Combining like terms, — 58;r = — 87. 

Changing signs, 58;r = 87; 

hence, x = ^ = 1Ans. 

{b) (^a^j^xY = x^ + 4.a‘^^a\ 

Performing the operation indicated in the first member, 
the equation becomes 

-j- = x^ + 

Canceling a* and x"^ (Art. 576), 

2^ V = 

Dividing by 2^®, x = 2. Ans. 

Clearing of fractions, the equation becomes 

{x — 1) (;ir -j- 2) — (;r -f 1) {x — 2) = 3. 

Expanding, x^'-^-x—2 — x'^-{-x-{-2 =3. 

Uniting terms, 2;r = 3. 

X = ^ = 1^, Ans. 

( 260 ) 159° 27' 34.6^ 

25° 16' 8,7" 

3° 48' 53. " 

188° 32' 36.3" 

( 261 ) (a) x+f + z-(x-j>)-{j> + z)-(-y) 
becomes x-{-j/-{-js — x-\-j/-~j/— on the removal of 
the parentheses. (Art. 482 .) 

Combining like terms, x — x + j J j j ^ ^ = 2j/. 

Ans. 

(6) {2x - y -jr 4:Z) + ( — X — j/ - iz) — (Sx — — z) 

becomes 2x — jf -j-iz — x — y — iz — 3x + 2j/ + z on the 





TRIGONOMETRIC FUNCTIONS. 


127 


removal of the parentheses. (Arts. 482 and 483.) Com¬ 
bining like terms, 

2;r — .IT — 3;r — j -j- 2/ + — 4^ -1- .S’ = .S' — 2.r. Ans. 

(c) a — [2a + {^a — 4^:)] — 5a— {Qa — [(7a 8«) — 9^] }. 

In this expression we find aggregation marks of different 
shapes, thus, [, (, and {. In such cases look for the cor¬ 
responding part (whatever may intervene), and all that is 
included between the two parts of each aggregation mark 
must be treated as directed by the sign before it (Arts. 482 
an d 483), no attention being given to any of the other 
aggregation marks. It is always best to begin with the 
innermost pair, and remove each pair of aggregation marks 
in order. First removing the parentheses, we have 
a — [2a -{-'da — 4«] — 5a — {6a — [7a 8 ^^ — 9 «]}. 

Removing the brackets, we have 

a — 2a — ?>a^a — 5a — {6a — 7a — Sa 9a}, 
Removing the brace, we haye 

a — 2a — da 4:a — 5a — 6a 7a Sa — 9a. 
Combining like terms, the result is — 5a. Ans. 


(262) (a) — 7my ) 35my -|- 2 dm^y" — 14?/^/* 

— 5m^ — ^my -j- 2y’^. Ans. 

(Art. 506.) 

(b) a*) 4a* — 3a^b — a^b"" 

4 — dab — a^b"^. Ans. 

(c) 4x^ ) 4x^ — + 12;r^ - 16x^ 

;ir — 2 x^ -f- dx'' — 4x\ Ans. 


( 263 ) Let X ■= the length of the post. 

X 

5 

3;ir 


Then, 


= the part in the earth. 


= the part in the water. 


From the conditions of the problem, we have therefore 
the following statement: 





1.28 


ELEMENTARY ALGEBRA AND 


from which 

and 

(264) 

Then, 


f + T + ” = -'‘ 

7;r+15;r+ 455 =: 35;ir; 

- 13a: = - 455; 
jtr = 35 feet. 


Let 

3 

;ir 


Ans. 

= the whole quantity. 

-]- 10 — the quantity of niter. 


— 4- = the quantity of sulphur. 

6 2 

1 I2x \ 

7 ~ ~ quantity of charcoal. 


Hence, 


*=T + >"+8 


-4+l(¥+“) 


Clearing of fractions and expanding terms, 

4:2x = 2Sx + 420 + 7;r - 189 + 4.^r + 60 - 84. 
Transposing, 

42;r - 2Sx ->7x- 4:X = 420 - 189 + 60 - 84. 

3;ir = 207. 

.;r = 69 lb., the quantity of gunpowder. 

Ans. 

2 V 69 

— 10 = — - -h lb = 56 lb., the quantity of niter. Ans. 

o O 

X 1 69 1 

- — 4z- — — -4-= 7 lb., the quantity of sulphur. Ans. 

6/16 /V 

56 lb. X T “ ^ = 6 lb., the quantity of charcoal. Ans. 


(265) (<^) See Art. 568. The cube root of — 125 is 

— 5. Dividing each of the exponents of the literal part by 3, 
the index of the root, the cube root of is x^y^s^ = xy^z ^; 

hence, \/ — 125x^y^2'' = — 5;rjV. Ans. 

{b) and (c) Proceed exactly as in {a). |/l0,000= ± 10 

and |/243 = 3; \/a'^b'''‘c'^= a^b^d — a^b^d^ and = m^'n\ 

Hence, ^ lOa^dd. Ans. 








TRIGONOMETRIC FUNCTIONS. 


129 


= ^in^n\ Ans. 

(d) Dividing the exponent of each letter in both numera¬ 
tor and denominator by 5, the index of the root, 






xy^z^ 

a^b^c^d' 


Ans. 


( 266 ) Using the proportion of Art. 615 , 

A B \ B C = sin C : sin 
or 70 : 42 = sin C : sin 36° 10'. 

Hence, sin (7 = X sin 36° 10' = ff x .59014 = .98357. 

The angle whose sine is .98357 is 79° 36'; hence,angle C ~ 
79° 36'. Ans. 

Angle B = 180° - {A A-C) = 180° - (36° 10' + 79° 36') = 
64° 14'. Ans. 

Using the proportion again, 

A C : B C = sin B : sin A ; 

or ^ C: : 42 ft. = sin 64° 14' : sin 36° 10' = .90057 : .59014 

TT ^ 42ft. X. 90057 . 

Hence, A C = -EH7ITI-“ 


( 267 ) (a) See Art. 591 . 

180° - 72° 11' 36" = 107° 48' 24". Ans. 

(b) See Art. 590 . 

90° - 22° 34' 17" = 67° 25' 43". Ans. 

( 268 ) Angle A= 180° - (yl + (S') = 

180° - (57° 34.5' + 44° 22.5') = '^'8° 3'. Ans. 

(S’ : ^ A = sin ^ : sin (S’ = sin 78° 3' : sin 44° 22.5' 
= .97833 : .69936. 


Hence, A C = 


344 ft. X .97833 
.69936 


= 481.22 ft. 


Ans. 


B C : A B=sin A : sin C= sin 57° 34.5' : sin 44° 22.5' 
= .8441 : .69936. 


Hence, B C = 


344 ft. X .8441 
.69936 


= 415.19 ft. 


Ans. 









*•^1 



0 


LOGARITHMS. 

(QUESTIONS 269-283.) 


( 269 ) First raise to the . 29078 power. Since = 2, 

( 200 \ 

_ \ ^2 and log 2 = . 29078 x log 2 = . 29078 X 

.30103 = .08753. Number corresponding = 1.2233. Then, 


1 - 



1.2283 = - .2233. 


We now find the product required by adding the log¬ 
arithms of 351.36, 100, 24, and .2233, paying no attention to 
the negative sign of .2233 until the product is found. (Art. 
647 .) 

Log 351.36 = 2.54575 
log 100 = 2 
log 24= 1.38021 
log .2233 = 1.34889 
sum = 5.27485 = 


r / 2 oo\ 

log 351.36 X 100 X 24 I 1 - j I 
Number corresponding = 188,300. 

The number is negative, since multiplying positive and 
negative signs gives negative; and the sign of .2233 is 
minus. Hence, 

X = — 188,300. Ans. 


(270) (^) Log 2,376 = 3.37585. Ans. (See Arts. 625 

and 627.) 

{b) Log .6413 =1.80706. Ans. 

{c) Log .0002507 = 4.39915. Ans. 

For notice of copyright, see page immediately following the title page. 




132 


LOGARITHMS. 


( 271 ) (a) Apply rule, Art. 652 . 

Log 755.4 = 2.87818 
log .00324= 3.510 55 

difference = 5.36763 = logarithrru of quotient. 

The mantissa is not found in tjie table. The next less 
mantissa is 36754. The difference between this and the 
next greater mantissa is 773 — 754 = 19, and the P. P. is 
763 — 754 = 9. Looking in the P. P. section for the column 
headed 19, we find opposite 9.5, 5, the fifth figure of the 
number. The fourth figure is 1, and the first three figures 
233; hence, the figures of the number are 23315. Since the 
characteristic is 5, 755.4 .00324 = 233,150. Ans. 

(b) Apply rule. Art. 652 . 

Log .05555 = 2.74468 
log .0008601 = 4.934 55 

difference = 1.81013 = logarithm of quotient. 

The number whose logarithm is 1.81013 equals 64.584. 

Hence, .05555 .0008601 = 64.584. Ans. 

(c) Apply rule, Art. 652 . 

Log 4.62 = .66464 
log .6448 = 1.80943 

difference = .85521 = logarithm of quotient. 

Number whose logarithm = .85521 = 7.1648. 

Hence, 4-62 .6448 = 7.1648. Ans. 


( 272 ) 


_ 238 X 1,000 
~ . 0042 '®®"" ■ 

Log 238 = 2.37658 
log 1,000 = ^_ 

sum = 5.37658 = log (238 X 1,000). 
Log .0042 = 3.62325 
.6602 
124650 
373950 
373950 


.411 469650 or .41147. 









LOGARITHMS. 


133 


.6G02 
- 3 

— 1.9806 = characteristic. 

Adding, .41147 
- 1.980 6 

2.43087 (See Art. 659.) 

Then, log = 5.37658 - 2.43087 = 6.94571 = 

6 94571 

.74 log x; whence, log .s; = —— = 9.38609. Number 
whose logarithm = 9.38609 is 2,432,700,000 = ;r. Ans. 

( 273 ) Log .00743 = 87099 

log .006 = 3.77815. 

= log .00743 H- 5 (Art. 662 ), and ‘l/dXie = 
log .006-^.6. Since these numbers are wholly decimal, 
we apply Art. 663 . 

5 )3.87099 

1.57419= log V-00743. 

The characteristic 3 will not contain 5. We then add 2 
to it, making 5. 5 is contained in 5, 1 times. Hence, the 

characteristic is 1. Adding the same number, 2, to the 
mantissa, we have 2.87099. 2.87099 5 = .57419. Hence, 

log Ko0743 = 1.57419. 

.6) 3.77815 .6 is contained in 3, — 5 times. 

5. .6 is contained in .77815, 1.29691 times, 

1.29 691 

sum = 4.2 9 6 9 1 = log 

Log i/Wm = 1.57419 
log'1/7^ = 4. 29691 

difference = 3.27728 = log of quotient. 

Number corresponding = 1,893.6. 

Hence, = 1,893.6. Ans. 














134 


LOGARITHMS. 


( 274 ) Apply rule, Art. 647 . 

Log 1,728 = 3.23754 
log .00024 = 4.38021 
log .7462 = 1.87286 
log 302.1 = 2.48015 
log 7.6094= .88135 
su7n = 2.S5211 = 

log (1,728X.00024X.7462x302.lx7.6094). Number whose 
logarithm is 2.85211 = 711.40, the product. Ans. 

( 275 ) Log 4 / 5 . 954 = .77481 2 = .38741 

log 4 / 6 I.I 9 = 1.78668 3 = .59556 

sum = .98297 
log |/^04 = 2.47500 5 = .49500. 

Then, 

/|/5.95 4 X ^ iQg ( 4 / 5.954 X |/61.19) - log 1/298.54 = 

4 / 298.54 

.98297 — .49500 = .48797 = logarithm of the required result. 
Number corresponding = 3.0759. Ans. 

( 276 ) V.0532864 = log .0532864 7. 

Log .0532864 = 2.72661. 

Adding 5 to characteristic 2 = 7. 

Adding 5 to mantissa = 5.72661. 

7 7 = L. 

5.72661 4- 7 = .81809, nearly. 

Hence, log 4 /. 0532864 = 1.81809. 

Number corresponding to log 1.81809 = .65780. Ans. 

( 277 ) (a) 32^•^ 1.50515 

Log 32 = 1.50515. _4^ 

1204120 

602060 

7.224720 

7.22472 is the logarithm of the required power. (Art. 

657 .) 

Number whose logarithm = 7.22472 is 16,777,000* 

Hence, 32^-®= 16,777,000. Ans. 


















LOGARITHMS. 


135 


{b) .76^-"’'^ 

Log .76 = 1.88081. 1 + .8 8081 

(See Arts. 658 and 659.) 3.6 2 

176162 

528486 

264243 

3.188 5 322 
-3.6 2 

1.56 85 3 = log .37028. 
Hence, .76"-” = .37028. Ans. 

{c) .84-+ 

Log .84 = 1.92428. 1+ .9 2 428 

.38 

739424 

277284 

.3512264 
-.3 8 

1.97 123 = log .93590. 

Hence, .84*’"* = .93590. Ans. 

e/' ] 

(278) Log y — log y = logarithm of answer. 

Log|/ii^ = i(logl - log 249) =-J-(0-2.39620)= -.39937 
= (adding + 1 and — 1) 1.60063. 

Log |/|| = I (log 23 - log 71) = \ (1.36173 - 1.85126) = 
|(_ .48953) = - .097906 = (adding + 1 and - 1) T.902094, 
or 1.90209 when using 5-place logarithms. 

Hence, 1.60063 - 1.90209 = 1.69854 = log .49950. 
Therefore, |/ff = .49950. Ans. 

(279) The mantissa is not found in the table. The next 
less mantissa is .81291 ; the difference between this and the 








13G 


LOGARITHMS. 


next greater mantissa is 298 — 291 — 7, and the P. P. is 
.81293 — .81291 = 2. Looking in the P. P. section for the 
column headed 7, we find opposite 2.1, 3, the fifth figure of 
the number; the fourth figure is 0, and the first three 
figures, 650. Hence, the number whose logarithm is .81293 
is .65003. Ans. 

2.52460 = logarithm of 334.65. Ans. (See Art. 640.) 

1.27631 logarithm of .18893. Ans. We choose 3 for 
the fifth figure because, in the proportional parts column 
headed 23, 6.9 is nearer 8 than 9.2. 

(280) The most expeditious way of solving this ex¬ 
ample is the following: 


p 




or — 



Substituting values given, = 1.495 


1.4l/ 

V 


134.7 

16.421' 


Log 7/j = log 1.495 


log 134.7 - log 16.421 
1.41 


.17464 + 


_j_ .64831 = .83285 = log 6.6504; 

1.41 

whence, i;, = 6.6504. Ans. 


(281) Log 




7.1895 X 4,764. S’’ X 0.00336“ 


i[log 7.1895 


.000489 X 457“ X .576’ 

+ 3 log 4,704.y- 5 log .00336 - (log .000489 + 3 log 457 + 


2 log .576)] 


5.77878 -4.1899 1 


log .030786. 


Ans. 


Log 7.1895 = .85670 

3 log 4,764.3 = 3 X 3.67799 = J7.35598 
5.log .00336 = 5 X 3.51323 = 13.56610 


S2im= 5.77878. 


Log .000489 = 4.68931 
3 log 457 = 3 x 3.65993 =_7.97976 
3 log .576 = 3 X 1.76043 =■!.53084 


sum = 4.18991. 












LOGARITHMS. 


13? 


(282) Substituting the values given, 


8,000 

X Vie; 
X 2.25 



2.25 


Log p = log 8,000 + 2.18 log j\ — log 2.25 = 3.90309 4 
2.18 (log 3 - log 16) - .35218 == 3.55091 + 2.18 X (.4?712 
- 1.20412) = 1.96605 = log 92.480. Ans. 


(283) Solving for A ^ =V 
Substituting values, given. 


/ ^ 

M xmx? = * 4 / 1)44 

V mm 
%m 
ms> 

1,000 

^ _ log .044 _ 2.64345 _ — 2.18 + .82345 
2.18 “ 2.18 “ 2.18 

1.37773 = log . 23863. Ans. 


ll.U. IV.—V 















r 



mu f 

li';?.; w,. * 


ro > '^*te 


' " t! * ' . 4.-> 

. V'^*-»'>4 ’ ' .'*»*' 'ro -- 

mM r. ' ' • 1 


- - m 


^r5£ ' 


•j . 




♦ - . 4r i y ^ 

V. > >i-j;-’i 

...^ . 


V ■/ - •-'vy- ■- 


.'S »;<fW’ ■ '5 


:'5-':- * - i I 

-- 



•‘•'Sk.' 




L?4;> 




‘A.,- 


‘ V -. 


4 5; 


zs^ 





K.»J1, 




*> ?7 


i^t: 








< ' 


4a 




Av''-:;>a:fi^ 

f'si'^V' • 

-'**>. .* 




a 


ii t-, q 




4? 


. *4 m: 


m^: 


►.*' .v'' 




?j -1 


"jci*-;; 


-'ll 




trA- 












“tl 










vk:' rl 




IAa 




1^. 


»>"i. 


y 


fi 


'•1»V 


I 


Si V" 




J ---“-w 


/■ •• - • 


,pl « 




p\^’%' 








• 








'J 


f>^' 




f*) 


•’i 












\i 




•^\L 




>tA; 


p- 


u.»'^‘..-.-ev-’ .. 1 


k 


p? WjU;iir/v} 




r %• 


ii* Ui 


*?3 




»srA< 


'3a- 




iW- ■ i V- ' 




m 




■>>k« 




t'A ; * 




<1 






' ' ^ - ‘ T L, '■ 

*'"-Tj 
' **? - ' 4(11' 

. .-1- 


^ v”\ ■ ‘ 




:.^^- 


V^ 


•.,— A’ 


k; ,>;rj ;-->>' 






> A 


ti 


r .■; 


* a» 


*. 














v;..'iV 


ST''--* .1 














I. • 






'U » 


'4#:' 


>/, '. 

►I * ■» 


.> 

« ^ --f^ 


4"' 


>. 


X • 


f” ^. 


«' 




7*;; 


r^ilv' 


>Si 




Si 


^01 










^•^’'•>:-v - . " } . > A^-', ' 






A . 


r«^ ^ tf- 





> *'* *' ' ^ ’■ ' 


r- 








*^*^,v J : ^ • m 


•’ '■' * • ■ ■ 

I <■*; f~ ■ r, \ f*T *v -^tL' V * 


;>5.--.v V-’IV 


' P 


fId ' "• • . 


» ' Ul 







ELEMENTARY MECHANICS. 

(QUESTIONS 355-453.) 


(355) Use formulas 18 and 8. 

Time it would take the ball to fall to the ground = t = 


2 h _ ./2 X 5.5 
g ^ 32.16 


.58484 sec. 


The space passed through by a body having a velocity of 
500 ft. per sec. in .58484 of a second = 5 = F ^ = 500 X 
.58484 = 292.42 ft.* Ans. 


(356) Use formula 7. 

||_X 3.1416 X.160 ^ 

oO 

(357) 160 60 X 7 = revolution in ^ sec. 360° X 

/ii 4 

A = 137^ = 137° 8' 34^ : Ans. 

L i I 

(368) {a) See Fig. 20. 36’ = 3'. 4 4- 3 = | = number 

o 

of revolutions of pulley to one revolution of fly-wheel. 

4 

54 X ^ = 72 revolutions of pulley and drum per min. 100 -r- 

O 

X 3.1416^ = 21.22 revolutions of drum to raise elevator 

01 OO 

100 ft. ■■ — X 60 = 17.68 sec. to travel 100 ft. Ans. 

I /V 

(b) 21.22 : 4.- :: 30 : 60, or x = = 42.44 rev, 

ou 


For notice of copyright, see page immediately following the title page. 








140 


ELEMENTARY MECHANICS. 


per min. of drum. The diameter of the pulley divided by 



Fig. 20. 


3 

the diameter of the fly-wheel = which multiplied by 
42.44= 31.83 revolutions per min. of fly-wheel. Ans. 

(359) See Arts. 857 and 859. 

(360) See Art. 861. 

(361) See Arts. 843 and 871. 

(362) See Art. 871. 

(363) See Arts. 842, 886, 887, etc. 

The relative weight of a body is found by comparing it 
with a given standard by means of the balance. The abso¬ 
lute weight is found by noting the pull which the body will 
exert on a spring balance. 

The absolute weight increases and decreases according to 
the laws of weight given in Art. 890 ; the relative weight is 
always the same. 

(364) See Art. 861. 

(365) See Art. 857. 














ELEMENTARY MECHANICS. 


141 


(366) See Art. 857. 

(367) If the mountain is at the same height above, and 
the valley at the same depth below sea-level respectively, 
it will weigh more at the bottom of the valley. 


Q1 

(368) -t-VoTT ~ ^ miles. Using formula 12, : AL: 

Oj /CoU 

M/ I, 3,900’X 20,000 

IV : tt/, we ha ve to = - j i - — ~— g -= 19,939.53 + lb. 

= 19,939 lb. si oz. Ans. 


(369) 

dlV 

oz. Ans. 


Using formula 
_ 3,958 X 20,000 
3,900 


11, : d :: W: Wj we have 

= 19,989.89 lb. = 19,989 lb. 14^ 

4 


(370) See Art. 870. 

(371) See Art. 894. 

(372) The velocity which a body may have at the in¬ 
stant the time begins to be reckoned. 


(373) Because the man after jumping tends to continue 
in motion with the same velocity as the «train, and the sud¬ 
den stoppage by the earth causes a shock, the severity of 
which varies with the velocity of the train. 


(374) See Arts. 870 and 871. 

(375) See Art. 872. 

(376) That force which will produce the same final 
effect upon a body as all the other forces acting together is 
called the resultant. 


(377) (a) If a 5-in. line = 20 lb., a 1-in. line = 4 lb. 

1 4 = T in. = 1 lb. Ans. (d) 6^ 4 = 1.5625 in. = 

4 4 

lb. Ans. 

4 

(378) Those forces by which a given force may be 







142 


ELEMENTARY MECHANICvS. 


replaced, and which will produce the same effect upon a 
body. 

(379) Southeast, in the direction of the diagonal of a 
square. See Fig. 21. 

(380) 4' 6' = 64". 64 X 3 XIX 

.261 = 21.141 lb. =: weight of lever. ^ 
Center of gravity of lever is in the 
middle, at Fig. 22, 27" from each 
end. Consider that the lever has no 
weight. The center of gravity of 
the two weights is at b, at a dis- 
47 X 54 

tance from c equal to — ^ = 21.508" = be. Formula 20, 

Art. 911. 




Consider both weights as concentrated at b, that is, 
imagine both weights removed and replaced by the dotted 
weight IF, equal to 71 -j- 17 = 118 lb. Consider the weight 
of the bar as concentrated at a, that is, as if replaced by 
a weight w = 21.141 lb. Then, the distance of the balancing 

point /, from or fe, = = • 834",' since ^ = 


27 — 21.508 = 5.492". Finaliy,/r + ^= fh = .834-f-21.508 
= 22.342" = the short arm. Ans. 54 — 22.342= 31.658" = 
long arm. Ans. 
















ELEMENTARY MECHANICS. 


143 


(381) See Fig. 23. 






(383) 46 — 27 = 19 lb., acting in the direction of the 

force of 46 lb. Ans. 










144 


ELEMENTARY MECHANICS. 


(384) {a) 18 X GO X GO == G4,800 miles per hour Ans. 

(^) G4,800 X 24 = 1,555,200 miles. Ans, 



Fig. 25. 





ELEMENTARY MECHANICS. 


145 


(385) («) 15 miles per hour = ^ == 

sec. As the other body is moving 11 ft. per sec., the 
distance between the two bodies in one second will be 
22 + 11 = 33 ft., and in 8 minutes the distance between 
them will be 33 X 60 X 8 = 15,840 ft., which, divided by the 

number of feet in one mile, gives = 3 miles. Ans. 

5,280 

(d) As the distance between the two bodies increases 
33 ft. per sec., then, 825 divided by 33 must be the time 

825 

required for the bodies to be 825 ft. apart, or = 25 
sec. Ans. 

(386) See Fig. 25. 

(387) (a) Although not so stated, the velocity is 
evidently considered with reference to a point on the shore. 
10 — 4 = 6 miles an hour. Ans. 

(d) 10 + 4 = 14 miles an hour. Ans. 

(c) 10 —4 + 3 = 9, and 10 + 4+3 = 17 miles an hour. Ans. 






146 


ELEMENTARY MECHANICS. 


(389) See Fig. 27. By relations I and II, Art. 609, 
be sin 23° = 87 X .39073 = 33.994 lb., ^ ^ = 87 cos 23° == 
87 X .92050 = 80.084 lb. 



Fig. 27. 


(390) See Fig. 28. (b) By relations I and II, Art. 

609, bc=S25 sin 15° = 325 X .25882 = 84.12 lb. Ans. 

(a) a e= 325 cos 15° = 325 X.96593 = 313.93 lb. Ans. 



Fig. 28. 

(391) TJse formula 10. 

IV 125 


MUf, 









ELEMENTARY MECHANICS. 


141 


(392) Using formula lO, >« = —, IV = mg=. 53.7 X 
33.16 = 1,727 lb. Ans. 


(393) (a) Yes. (d) 35. (c) 25. Ans. 

(394) (a) Using formula 12, cC •. R' :: W w, d = 


^ |/li000^xlH ^ 4 749.736 miles. 4,749.736 - 
w 100 

4,000 = 749.736 miles. Ans. 


(^) Using formula 11 , R : d :: W : w, d = 


R w 

~W 


4,000 X 100 
141 

miles. Ans 


2,836.88 miles. 4,000 - 2,836.88 = 1,163.12 


(395) (a) Use formula 18, 


t 



2 X 5,280 
32.16 


= 18.12 sec. 


Ans. 


(b) Use formula \ — gt — 32.16 X 18.12 = 582.74 ft. per 

sec., or, by formula 16, = 4/2 X 32.16 X 5,280 = 

582.76 ft. per sec. Ans. 

The slight difference in the two velocities is caused by not 
calculating the time to a sufficient number of decimal places, 
the actual value for t being 18.12065 sec. 

Wv^ 

(396) Use formula 25. Kinetic energy = Wk = ■. 

Wh = 160 X 5,380 = 844,800 ft.-lb. 


IVV _ 160 X 582.76’ 
2g ~ 3 X 33.16 


844,799 ft.-lb. Ans. 


(397) (a) Using formulas 15 and 14, 

= I? = 

t = —= time required to go up or fall back. Hence, total 
S' 

2v 2 X 2,360 o 44 ^-, • o • . 70 . 

time = — sec. = — = 2.4461 min. = 2 mm. 26.77 

g bO X o2.1b 

sec. Ans. 















* 


U 8 ELEMENTARY MECHANICS. 

(398) 1 hour = 00 min., 1 day = 24 hours; hence, 1 
day = CO X 24 = 1,440 min. Using formula 7, V = -; 

whence, = 17.453 + miles per min. Ans. 

i,44U 

(399) {a) Use formula 25. 

... . 400 X 1,875 X 1,875 

Kinetic energy = —— =- 2 X 32 16 ~^-” 

ft.-lb. Ans, 

(^) ^1.803 339.55 ^ 

yV j UUU 

(c) See Art. 961. 

Striking force X 4 = 21,863,339.55 ft.-lb., 

iZ 


or striking force 


21,863 339.55 _ 43 p,. 


Ans. 


(400) Using formula 18, t = i/— 
3.52673 sec., when^= 32.16. 


i/^ 


2 X 200 _ 


32.16 


t = X 200 _ ^ 47214 sec., when g = 20 . 

^ 20 

4.47214 - 3.52673 = 0.94541 sec. Ans. 

(401) See Art. 910. 

(402) See Art. 963. 

(403) (a) See Art. 962. 

800 r. W 500 


^ 


V 


Hence, D 


gv 1,728- ’ gv 32.16XA-V-S 

33.582. Ans. {b) In Art. 962, the density of water was 
found to be 1.941. (f:) In Art. 963, it is stated that the 

specific gravity of a body is the ratio of its density to the 

33 582 

density of water. Hence, = 17.3 = specific gravity. 

If the weight of water be taken as 62.5 lb. per cu. ft., the 
specific gravity will be found to be 17.28. Ans. 

(404) Assuming that it started from a state of rest, 
formula 13 gives 2 ^ = ^ = 32.16 X 5 = 160.8 ft. per sec. 














ELEMENTARY MECHANICS. 


149 


(405) Use formulas 17 and 13, 



3/3 ] 6 

—^ - X 3’ = 144.72 ft., distance fallen at the end of third 
second. 

V = g‘f = 32.1(3 X 3 = 9().48 ft. per sec., velocity at end of 
third second. 


9(3.48 X 6 = 578.88 ft., distance fallen during the remain¬ 
ing 6 seconds. 

144.72 + 578.88 = 723.6 ft. = total distance. Ans. 


(406) See Art. 961. 

Striking force x = 8 X 8 = 64. Therefore, striking 

force = = 1,536 tons. Ans. 

X 

12 

(407) See Arts. 901 and 902. 

(408) Use formula 1 9. 

Centrifugal force = tension of string = .00034 WR = 

.00034 X (.5236 X 4* X .261) x ^ X 60’ = 13.38+ lb. Ans. 

±z 


(409) (SO’* - 70") X .7854 X 26 X .261 2 = weight of 

^ of rim. 

^ 80 - 10 35 . 

^ = Wi2 =12^‘- 

According to Art. 904, F= .00034 JVR N"" -j- 3.1416 = 

.00034 X X i X 175’ 

Z lZ 

3.1416 = 38,641 lb. Ans. 

(410) (a) Use formulas 11 and 12. R : d :: IV: 

wA 1 x 4,000 .... . 

or IV = -T- = --= 40 lb. Ans. 

a 100 

4 000*^ V 40 

(d) : A" :: IV : w, or ’ J:, --- = 38.072 lb. Ans. 


(411) See Art. 955. 
10,746 X 354 


10 X 33.000 


4,100" 

11.5275 H. P. Ans. 








150 


ELEMENTARY MECHANICS. 


(412) Use formula 12. 

d' : R'wW-.w, or d= = i3_064 mi., nearly, 

13,064 — 4,000 = 9,064 miles. Ans. 

(413) Use formula 18. t = ~ 

1.7634 sec., nearly. 

1.7634 X 140 = 246.876 ft. Ans. 

(414) ^ i sec. Use formula 17. 

h = = |- X 32.16 X = 1.78| ft. = 1 ft. 9.44 in. 

' Ans. 



(415) See Arts. 906, 907. 

(416) See Arts. 908, 909. 










ELEMENTARY MECHANICS. 


151 


(417) No. It can only be counteracted by another 
equal couple which tends to revolve the body in an opposite 
direction. 

(418) See Art. 914. 

(419) Draw the quadrilateral as shown in Fig. 29. 
Divide it into two triangles by the diagonal B D. The 
center of gravity of the triangle B C D found to be at a, 
and the center of gravity of the triangle A B D found to 
be at b (Art. 914). Join a and b by the line a b, which, on 
being measured, is found to have a length of 4.27 inches. 
From C and A drop the perpendiculars C F and A G on the 

diagonal B D. Then, area of the triangle A B D = \-{AG 

/i 

B B), and area of the triangle B C FylB D). 

/V 

Measuring these distances, B Z> = 11', C /'= 5.1', and 
A G=1l.V. 

Area A BD = \ X 7.7 X 11 = 42.35 sq. in. 

Area ^ ^ x 5.1 X H = 28.05 sq. in. 

2 

According to formula 20, the distance of (9, the center of 
gravity, from b is - - —ri W ' ok ~ Therefore, the cen- 

UO “p 4:yC.O0 

ter of gravity is on the line ab 2 X 2 i. distance of 1.7'" from b. 

(420) See Fig. 30. The 
center of gravity lies at the 
geometrical center of the penta¬ 
gon, which may be found as 
follows: From any vertex draw 
a line to the middle point of the 
opposite side. Repeat the 
operation for any other vertex, 
and the intersection of the two 
lines will be the desired center 
of gravity. 



Fig. 30. 




u% 


ELEMENTARY MECHANICS. 


(421) See Fig. 31. Since any number of quadrilaterals 
can be drawn with the sides given, any number of answers 
can be obtained. 

Draw a quadrilateral, the lengths of whose sides are equal 
to the distances between the weights, and locate a weight on 
each corner. Apply formula 20 to find the distance ; 


Q V 18 

thus, Wj = = 5.4". Measure the distance 

y /wJ. 



suppose it equals say 36". Apply the formula again. 
(^1 M^^sure W^; it equals say 

31.7". 

Apply the formula again. C.C= ^ -=8 7" 

^ ^ ^ 17-pi5-p9_|_21 

C is center of gravity of the combination. 

(422) Let A B C D Fig. 32, be the outline, the 
right-angled triangle cut-off being E S D. Divide the 
figure into two parts by the line in w, which is so drawn 
that it cuts off an isosceles right-angled triangle m B 
equal in area to E S from the opposite corner of the 
square. 






ELEMENTARY MECHANICS. 


153 


The center of gravity of A mn CDE is then at , its 
geometrical center, = 4 in.; angle Bmr — 45°; there- 



center of gravity of Bmn^ lies on Br^ and BC^=:—Br 

O 

= 1x2.828 = 1.885 in. £C^ = ABxsm BAC^=Ux 

o 

sin 45° = 14 X .707 = 9.898 in. C^C^ = B - B C, = 9.898 
- 1.885 = 8.013 in. 

Area ABCDE=\^ — - = 188 sq. in. Area inBn 

_ _ g Area A mn CEE = 1S8 — S = ISO 

2 

sq. in. 

The center of gravity of the combined area lies at C, at 


B.Q. IV.—10 










154 


ELEMENTARY MECHANICS. 


a distance from , according to formula 20 (Art. 911 ), 


equal to 
BC 


8 X 8.013 


180 + 8 188 
BC - 9.898 -.341 


= .341 in. C^C = .341 in. 
9.557 inches. Ans. 


( 423 ) {b) In one revolution the power will have moved 
through a distance of 2 X 15 X 3.1416 = 94.248", and the 

1 " 

weight will have been lifted j , The velocity ratio is then 

94.348^-1 = 376.993. 

4 

376.993 X 35 = 9,434.8 1b. Ans. 

{a) 9,434.8 - 5,000 = 4,434.8 lb. Ans. 

{c) 4,434.8 9,434.8 = 46.95^. Ans. 

( 424 ) See Arts. 920 and 922 . 

( 425 ) Construct the prism A B E Fig. 33. From 
A, draw the line E F. Find the center of gravity of the 



rectangle, which is at and that of the triangle, which is 






ELEMENTARY MECHANICS. 


155 


at C^. Connect these centers of gravity by the straight line 
C^ and find the common center of gravity of the body by 
the rule to be at C. Having found this center, draw the 
line of direction C G. If this line falls within the base, the 
body will stand, and if it falls without, it will fall. 

(426) {a) 5 ft. 6 in. = 66^ 66 6 = 11 = velocity 

ratio. Ans. 

{b) 5 X 11 = 55 lb. Ans. 


( 427 ) 55 X .65 = 35.75 lb. Ans. 

( 428 ) Apply formula 20 . 5 ft. = 60^ = 

9.7674 in., nearly, = distance from the large weight. Ans. 


( 429 ) (^) 1,000-4- 50 = 20, velocity ratio. Ans. See 

Art. 945 . {b) 10 fixed and 10 movable. Ans. (c) 50 -f- 95 

= 52.63^. Ans. 

( 430 ) Px circumference = X or 60 X 40 X 3.1416 
= Wxl, or IV= 60 X 40 X 3.1416 X 8 = 60,318.72 Ib. Since 

O 

the efficiency of combination is 40^, the tension on the stud 
would be .40 X 60,318.72 = 24,127.488 lb. Ans. 

( 431 ) (a) |/20’‘ + 5^ = 20.616 ft. = length of inclined 
plane. 

PX length of plane = JV X height, or Px 20.616 = 
1,580 X 5. 

P= = 383.2 lb. Ans. (/;) In the second case, 

20. bib 

P X length of base = JV X height, or Px 20 = 1,580 X 5; 

1,580 X 5 onpr IV. A 

hence, P= - ^ -= 395 lb. Ans. 

Z\) 

( 432 ) fFx2 = 42x6, or 1F=^^^^ = 126 lb. 

126 + 42 = 168 lb. 168 X 1 = 14^' X 12, or W' = ^= 14 lb. 

Ans. 







156 


ELEMENTARY MECHANICS. 


(433) See Fig. 34. Px 14 x 21 X 19 = 2^ X X 2| 
X 725, or 


P=. ^ = 3.032 lb. Ans. 

14 X 21 X 19 

IB' 


W 


— 24 ^—± 6 '^* 6 




- 30 ^^ 


4= 


18 ^ 


W 


(434) See Fig. 35. {a) 35 X 15 X 12 X 20 = 5 X S 

3 X or 

jj/- _ 35 X 15 X 12 X 20 

' 5 X 3» X 3 -=2,400 lb. Ans. 




^1 M lb. 



























ELEMENTARY MECHANICS. 


157 


4 

(d) 2,400 35 = 68y = velocity ratio. Ans. 

(c) 1,932 -r- 2,400 = .805 = 80.5^. Ans. 


(435) In Fig. 36, let the 12-lb. weight be placed at A, 
the 18-lb. weight at B, and the 15-lb. weight at B. 

Use formula 20. 



gravity of the 12 and 18-lb. weights from B. Drawing U, B, 

r r— N ^ = ^ c B. Measuring the distances 
^ (12 -j-18) -|- 15 3 ^ 

of C from BD, DA, and A B, it is found that Ca = 3.45", 
Cb = 5.25”, and Cd=^.V'. Ans. 


(436) (a) Potential energy equals the work which the 

body would do in falling to the ground = 500 X 75 = 
37,500 ft.-lb. Ans. 






158 


elementary mechanics. 


{d) Using formula 18, 


sec. = .035995 min., the time of falling. 


, ^-r ^ -r , 


X 75 
3^.16 


= 2.1597 


37,500 


= 31.57 H. P. Ans. 


33,000 X .035995 
(437) 12762.5 = 2.032 = specific gravity. Ans. 


(438) X 9.823 = .35529 lb. Ans. 

1,7 2o 


(439) Use formula 21. lU = ^ 

or ^ 0 ^ 5 ^^ 5 ^ 5 ^ X .48 = 499.2 lb. Ans. See ^ 
Fig. 37. 


- 

MW 


Ax 


(440) See Art. 961. 

/3 Wv^ 1.5X25’ 


2 X 32.16’ 


1.5 X 25’ 


^ 2x32.16 . 

F=— - -^= 466.42 lb. Ans. 

f 12 

(441 ) {a) 2,000 -f- 4 = 500 = wt. of cu. 

ft. 500 -f- 62.5 = 8 = specific gravity. Ans. 


500 



.28935 lb. Ans. 



Fig. 37. 


(442) See Fig. 38. 14.5 X 2 

= 29. 30 X 29 = fU X 5, or = 

30 X 29 . 

-z-= 174 lb. Ans. 


(443) 75 X.21 = 15.75 lb. 

Fig. 38 . Ans. 

(444) {a) 900 x 150 = 135,000 ft.-lb. Ans. 

135,000 „ „ , „ 

—— = 9,000 ft.-lb. per mm. Ans. 

,,, 9,000 3 „ , 

^ ^ 33,000 11 




































ELEMENTARY MECHANICS. 


159 


(445) 900 X .18 X 2 = 324 lb. = force required to over¬ 

come the friction. 900 + 324 = 1,224 lb. = total force. 


1,224 X 150 
15 X 33,000 


.37091 H.P. Ans. 


(446) 18 -f- 88 = . 2045. Ans. 


(447) SeeArt.962. =i?= = 12.438. 


(448) See Fig. 39. 125 

- 47.5 = 77.5 lb. = down¬ 
ward pressure. 

77.5^ 4 = 19.375 lb. 

= pressure on each support. 

Ans. 



Fig. 39. 


(449) See Fig. 40. 



(460) See Fig. 41. 4.5 2 = 2.25. 

1 o 

^ X 6 X 30 = 960 lb. Ans. 

Z. Zo 

(451) (a) 960-5- 30 = 33. Ans. 

(6) 790 -5- 960 = . 8339 = 83.39^. Ans. 









100 


ELEMENTARY MECHANICS. 


(452) (a) See Fig. 43. 475 + (475 X .34) = 589 lb. 

Ans 

{6) 475 589 = . 8064 = 80.64^. Ans. 



Pig. 41. 



Fig. 43. 


(453) (a) By formula 23, U = FS — 6 X 25 = 150 

foot-pounds. Ans. 

CM 34 1 . 

(^) 2^sec. = -=-mm. 

Using formula 24, Power = ^ = 3,600 ft.-lb. per 

mm. Ans. 













Pneumatics, Gas, and Petroleum. 

(QUESTIONS 504-658.) 


(504) ‘The force with which a confined gas presses 
against the walls of the vessel which contains it. (Art 

1040.) 

(505) (a) See Art. 1042. 4 X 12 X .49 = 23.52 lb. 

per sq. in. Ans. 

((^) 23.52 ^ 14.7 = 1.6 atmospheres. Ans.‘ 

(506) {a) A column of water 1 ft. high exerts a pres¬ 
sure of .434 lb. per sq. in. Hence, .434 x 19 = 8.246 lb. per 
sq. in., the required tension. A column of mercury 1 in. 
high exerts a pressure of .49 lb:* per sq. in. Hence, 8.246 -i- 
.49 = 16.828 in. = height of the mercury column. Ans. 
(Art. 1043.) 

(3) Pressure above the mercury = 14.7 — 8.246 = 6.454 lb. 
per sq. in. Ans. (Art. 1043.) 


(507) Using formula 53, 

^ (14.7X3) XI IK • A 

A = -= 17.64 lb. per sq. in. Ans. 

'T, 2.5 


(508) (a) 7.14-^.08 = 89.25 cu. ft., the new volume. 

Ans. 

(3) If 1 cu. ft. weighs .08 lb., 1 lb. contains 1-h .08 = 
12.5 cu. ft. Hence, using formula 60, 


/ V- .37052 r, or ^ 


22.05 X 12.5 
.37052 


743.887. 


The new temperature is therefore 743.887°— 460° = 283.887°. 

Ans. 


For notice of copyright, see page immediately following the title page. 





162 pneumatics; gas, and petroleum. 


(c) Using formula 61 , 


V = 


.37052 IV T .37052 X 7.14 X 535 


= 64.188 cu. ft. Ans. 


/ 22.05 

(r = 460°+ 75° = 535°, and /= 14.7 X 1.5 = 22.05 lb. per 

sq. in.) 

( 509 ) Substituting in formula 59 , 

/ 40, / = 120, and = 55, 

40 X 515 


^ V460 + 120/ 


580 


35.517 lb. persq. in. Ans, 


( 510 ) Using formula 61 , / F= .37052 IV 7", 


W: 


pV 


Therefore, W 


.37052 T' 
r= 460°+ 60° = 520°. 
14.7 X 1 


.37052 X 520 


= .076296 lb. Ans. 


(511) 175+00^ 144 = lb. per sq. in. (175,000-^-144)-^ 
14.7 = 82.672 atmospheres. Ans. 

(512) Extending formula 63 to include three gases, 
we have 

PV=p, ^^i+A^'a+A oi" 40x/^= 1 X12 + 2X10 + 3X8. 

Hence, 1.4 atmospheres = 1.4 X 14.7 = 20.58 lb. 

per sq. in. Ans. 

(513) In the last example, A* F= 1.4 X 40 = 56, taking 
the pressure in atmospheres. In the present case, P = 

23 

atmospheres. 

Therefore, U= ^ =-|^ = 35.79 cu. ft. Ans. 

147 

( 514 ) When / = 280°, T = 740° ; when / = 77°, T = 537°, 

j> V=. 37052 WT, or (Formula 61.) 









PNEUMATICS, GAS, AND PETROLEUM. 163 


Weight of hot air = 536.13 lb. 

w • u. c • 1 .5 14.7 X 10,000 „„„ ,, 

Weight of air displaced = —■ — —— = 738.81 lb. 

.o7U0/& X ooi 

738.81-536.13 = 202.68. 202.68-100 = 102.68 lb. Ans. 


(515) According to formula 64, PV — 
<14.7X13. 30 X 18^ 


or 20 X 


31 = ( 


533 


513 


20 V 31 

Therefore, T = — 4'39.35. Since this is less than 

1.411100 

460, the temperature is 460° —439.35° = 20.65° below zero, 
or-20.65°. Ans. 


(516) A hollow space from which all air or other gas 
(or gaseous vapor) has been removed. An example would 
be the space above the mercury in -a barometer. (Art. 

1043.) 


( 517 ) One inch of mercury corresponds to a pressure 
of .49 lb. per sq. in. 


inch of mercury corresponds to a pressure of lb. 

per sq. in. 

49 

^ X 144 = 1.764 lb. per sq. ft. Ans. (Art. 1042.) 

(518) 8.47 cu. ft. = original volume = 8.47 cu. ft. 

— 4.5 cu. ft. = 3.97 cu. ft. = new volume = v. By formula 53, 




3.97 X 38 
“ 8.47 


= 17.812 lb. per sq. in. 


Ans. 


(519) Applying formula 58, 



/460 + /A 

4,516, 

/ 460 + 80 \ 

\4t60 + t ) 

' 1,728' 

UhO + 260/ 


= 1.96 cu. ft. 


Ans. 


( 520 ) 48 in., 36 in., and 24 in. = 4 ft., 3 ft., and 2 ft., 

respectively. Hence, 4 X 3 X 2 = 24 cu. ft. = the volume 








1G4 PNEUMATICS, GAS, AND PETROLEUM. 


of the block. The block will weigh as much more in a 
vacuum as the weight of the air it displaces. In example 
510, it was found that 1 cu. ft. of air at a temperature of 
60° weighed .076296 lb. .076296 X 24 + 1,200 = 1,201.83 lb. 

Ans. 

(521) ^ {a) See Art. 1043. 

3Q_23 

(b) Pressure in condenser ——— X^14.7 = 3.43 lb. per 

oU 

sq. in. Ans. 

(522) 144 X 14.7 = 2,116.8 lb. per sq. ft. Ans. 


(523) .27 3 = .09 = weight of 1 cu. ft. 

Using formula 56, 

P Px IV^ = 65 X .09. = 0.195 lb. Ans. 

(524) Using formula 61, 

/ F= .37052 lUr, or 30 X 1 = .37052 X .09 X T. 

30 

7’== 899.6. 899.6°-460° = 439.6°. Ans. 


(525) 460° + 33° = 493°; 460° + 313° = 673°; 460° + 

63? = 533°; and 460° + (-40°) = 430°. 


(526) Using formula 61, / U= .37063 IVT, and sub¬ 
stituting, (14.7 X 10) X 4 = .37053 X 3.5 X T, 

14 7 X 10 X 4 

T^ 053 x 3.5 = 453.417°-460° = -6.583°. 

Ans. 

(527) Using formula 59, 


Px--=P 


/460 + 

Ueo +1 


) 


= 12 X 


/460 + 300\ 
\ 460 + 60 / 


= 17.54 lb. 


per sq. in. 
Ans. 


(528) 460°-j-212° = 672°. Using formula 61, 

p V= .37053 W r, we have 14.7 X 1 = .37053 X fUx 673, 


or 


IV = 


14.7 


.37052 X 672 


= .059039 lb. Ans. 








PNEUMATICS, GAS, AND PETROLEUM. 105 


(529) Using formula 63, 

VP= vp-\-vJ^, or 30 X 35 = 19 X 12 + 21/,, 


or 




30 X 35-19 X 12 
21 


39.14 lb. per sq. 


in. 


Ans. 


(530) (a) See Art. 1043. 

(d) The gauge will show a height of 30 in. — 4.5 in. = 
25.5 in. Ans. 


(531) Sp. Gr. of alcohol = .8, Therefore, 16X .434x .8 = 


pressure exerted by the column of alcohol. 


16X.434X.8 

.49 


11.337 in. = height of a column of mercury that will give 
the same pressure as 16 ft. of alcohol = number of inches 
shown by the gauge. Ans. 


(532) Using formula 56, 

pW^=p^ W, or 3.5 X 14.7 X 2=/, X 13; 

, ^ 14.7 X 3.5 X 2 . . 

hence, /, =-—-= 7.915+ lb. per sq. in. Ans. 

lo 

(533) 60 in. — 50 in. = 10 in. Since the volumes are 
proportional to the lengths of the spaces between the piston 
and the end of the stroke, we may apply formula 62, 

U, 14.7 X 60 _ /, X 10 
r - : or + 60 “ 460 + 130' 

f ^ 14.7 X 60 X 590 . . 

Therefore, /, =-—r---r-= 100.07 lb. per sq. in. Ans. 

520 X 10 


(534) T = 137“ + 460“ = 587“. Using formula 60, 


/ K = .37053 T, or U= = 8.055 cu. ft. Ans. 

(536) T = 100 “ + 460“ = 560“. 

Substituting in formula 61, / F = .37052 W 7", 


or F = 


.37052 WT 

P 


.37052 X .5 X 560 
4,000 


3.735 cu. ft. 


An§. 


144 












166 PNEUMATICS, GAS, AND PETROLEUM. 


(536) Use formula 64, PV — 


r = 110° + 460° = 570°; T, = 100° + 460' 
7; = 130° + 460 = 590°. 

/90 X 40 


560' 


Therefore, V = 


560 


?22il!'i57o 

590 


120 


(537) Using formula 58, 

'460 + 115 


Ueo + f;/ \ 


f) 


= 67.248 cu. ft. Ans, 


4.6 cu. ft. Ans. 


460 + 40 

(538) {a) A brownish-black substance resembling thick 
molasses, usually of a tarry odor. (Art. 1064.) 

(d) A combination of carbon and hydrogen. (Art. 1065.) 

(539) {a) Because a narrow belt follows a fissure or 
crevice in the rock. 

(d) Because a wide stretch of oil territory occurs over a 
level, or nearly level, oil bed. (Art. 1067.) 


(540) (a) The softer parts of animal remains. 

(I?) The occurrence of oil mixed with fossil bones, and 
the impossibility of its coming from outside sources, in 
many places where it is found. (Art. 1068.) 

(541) See Art. 1069. 

(542) (a) By barrels, tank-carts and cars, and pipe¬ 
lines. 


(d) Pipe-lines. (Art. 1070.) 

(543) {a) By distillation. (Art. 1071.) 

(d) Any three of the following: ether, gasoline, petroleum, 
naphtha, kerosene, lubricating oil, coke. (Art. 1073.) 

(544) (a) First mixing the distillate with acid in a large 
vat, stirring with air pressure. Then allowing the tarry 
acid to settle and drawing it off. Washing the distillate 
with water and finally running it into the lye-agitator. 
Stirring it with lye in the lye-agitator. 








PNEUMATICS, GAS, AND PETROLEUM. 167 


{d) Bymeansof settling tanks or by filtering. (Art. 1075.) 

(545) (a) Heating soft coal in a retort at a red heat. 
(Art. 1077.) 

(d) Because the tar in the hydraulic main seals the end 
of the pipe from the retort; or, the gas is pumped out by 
the exhauster. (Art. 1078.) 

(546) (a) A large box filled with wet coke or brush. 

(d) It removes the ammonia and tar. (Art. 1078.) 

(547) . To remove the sulphur compounds. (Art. 1078.) 

(548) {a) Anthracite coal and steam, or coke and 
steam. (Art. 1079.) 

(^) See Art. 1081. 

(549) (^) The products of incomplete combustion of 
coal. 

(d) Nitrogen. 

(c) No. 

(d) Because it will neither burn nor assist combustion, 
and carries away heat. (Art. 1082.) 

(550) (^) A mixture of producer and water gas. 

(d) Because it contains less nitrogen and more gas that 
will burn. 

(c) Because the process is continuous. (Art. 1083.) 

(551) (^) Oils of all kinds: petroleum, animal fats, 
and vegetable oils. (Art. 1084.) 

(I?) Lighting railway trains. (Art. 1085.) 

(552) (a) In the ground and generally in or near the 
oil regions. 

(d) Marsh gas. (Art. 1086.) 

(553) (^) A gas smelling like garlic, which burns with 
a very white flame. 

(^) By the action of water on calcium carbide. 

(c) Because of the present high price of the carbide. 
(Art. 1087.) 




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HEAT. 

(QUESTIONS 554-6ia) 


(554) 9 hr. 25 min. = 565 min. 

55 -7- 2 = 27.5 min. = time it would take to heat 1 lb. of 
the substance to this temperature under the same conditions. 
27.5 -7- 565 = .04867 = specific heat of the substance. Ans. 


(555) Increase in diameter = 4.001 ■— 3.9985 = .0025^ 
= .00000599, from Table 19. Hence, using formula 


. 1 -0025 

~ Z C, “ 3.9985 X .00000599 “ 
104.4 + 80 = 184.4“, nearly. Ans. 


Then, 


(556) {a) Substituting in formula 72 W = 360, — 
/ = 104.4, and J = .1165; n = IV (/,-/) .? = 360 X 104.4 
X .1165 = 4,378.536 B. T. U. 

Since 12^ of the total heat is lost by radiation, the above 
number of B. T. U. must be the number remaining after 
subtracting the heat units lost by radiation from the total 
number; or it is 100 — 12 = 88^ of the total number. 
Hence, dividing by .88 we obtain 4,378.536 .88 = 4,975.6 

B. T. U., total heat required. Ans. 

(d) 4,975.6 X 778 = 3,871,017 ft.-lb. Ans. 

(557) (a) 80“ X .7854 = 5,026.56 sq. in. = area of 

piston. 

p V = p, v^. At the beginning of the stroke p = 14.7, and 
the volume may be represented by the length of the cylinder, 
80 in.; at the point of discharge /, is 120, therefore, 80 X 14.7 
= 120 X or z/ = 9.8 in., the portion of the stroke uncom¬ 
pleted at discharge. 

5,026.56 X 9.8 __ 2g 507 cu. ft., volume of air discharged. 

An& 


For notice of copyright, see page immediately following the title page. 


E.G. IV.—11 





170 


HEAT. 


{b) By formula 80, L = 331.5744 p Flog ^ = 331.5744 X 

L20 X 28.507 log ^ = 331.5744 X 120 X 28.507 X .91186 = 
1,034,289 ft.-lb. Ans. 


(558) {a) Volume of cylinder = area of piston X length 

, , 5,026.56 80 , 

of stroke = ’ ,, — X rr; = 232.71 cu. ft., nearly. 

144 12 


Prom formula 81,/,^^,**^* = 


Pt 

Substituting for and their values at the beginning 
of the stroke, and for its value at discharge, we have 




14.7 X 232.71^'*^ 


120 


, log 14.7 + 1.41 log 232.71 - log 120 

log ^-, or 

, 1.16732 + 1.41 X 2.36682 - 2.07918 , 

log 2 /, =--k- — - = 1.72011. 

= 52.494 cu. ft., the volume discharged. Ans. 

(b) To find the work, use formula 84, 

Z = 351.36/= 

351.36 X 14.7 X 332.71 X [l - 

( 232 71 \ 

232.71 - log 53.494) = 

.41 (2.36683 - 1.72011) = .26515. Therefore, (||^)'“ = 
1.8414. 

1 — 1.8414 = — .8414, the minus sign indicating that the 
work done by the air is negative; that is, that work is done 
upon the air instead of by it, or that the air is compressed. 

L = 351.36 X 14.7 X 232.71 X .8414= 1,011,317 ft.-lb. 

Ans. 

(559) See Arts. 1090 and 1135. 












HEAT. 


171 


( 560 ) Using formula 65 , 

{a) f X 2,917 + 32 = 5,282.6° F. Ans. 

(d) />= I X 637 + 32 = 1,178.6 F. Ans. 

(c) » X (- 260) + 32 = - 436° F. Ans. 

( 561 ) See,Table 21, Art. 1134 , and Table 22, Art. 

1141 . 

- 50 - ( -37.8) X 12 X .0333 = 4.875 = units of heat 
necessary to raise the temperature to the fusion point. 
5.09 X 12 = 61.08 B. T. U. required to melt the mercury. 
[662 - ( - 37.8) ] X .0333 X 12 = 279.64 B. T. U. required 
to raise the temperature to the point of vaporization. 
157 X 12 = 1,884 B. T. U. required for vaporizing it. 
4,875 + 61.08 + 279.64 + 1,884 = 2,229.595 B. T. U. Ans. 


(502) Using formula 71,/ F= IVR T, or 


vv 


18 X 100 


5.34946 X 540 


= .6231 lb. 


Ans. 


(563) See Tables 21 and 22 for specific heat and tem¬ 
perature of fusion of copper. 7 lb. 5 oz. = 7.3125 lb. 
2,100° - 78° = 2,022°. 2,022 X 7.3125 X .0951 = 1,406.13 


B. T. U. 
(564) 


Ans. 

By formula 73, = 




2.25 X .0314 X 40 + 4 X 65d 1.75 X .1298 X 62 
2.25 X .0314+4 + 1.75 X .1298 


64.43°. 


Ans. 


(565) Use formulas 69, 67, and 68. {a) v=VC^t — 
4 X 4 X 20 X 1,200 X .00002058 = 7.903 cu. in. Ans. 

(b) 1 = LCJ = 20 X 1,200 X.00000686 = .16464 in. Ans. 

(c) a = ACj = 4.X 1,200 X .00001372 = .2634 sq. in. 

An.s. 

(566) Using formula 71, 

pV=WRT.o,T=-^ = liSSs = 

1,167.45° - 460° = 707.45°. Ans- 

(567) See Art. 1124. 









172 


HEAT. 


(568) By formula 73, t = 

-^1 + -^a H- -^3 

, _ + Ks,) IV,S, t, _ 

IV. s. 


or 


(1 X .1298 + 3.25 X 1 + 1.5 X -0951) 128 - 3.25 X 85 - 1.5 X .0951 X 85 


lx. 1298 


1,252°. Ans. 


(569) Draw the lines O X and O V (Fig. 46) perpen¬ 
dicular to each other. On O X take any convenient dis¬ 
tance, say ^ inch, to represent one unit of volume (1 cu. ft.), 
and onO Y take a distance of say 3 inches to represent the 
required pressure of 84 pounds. Our scale of pressure is, 



then, 28 pounds to the inch. The ordinates are now erected 
as shown in Art. 1156# Thus, at A the volume is 1, and 




































HEAT. 


173 


the pressure 84 pounds; hence, / = 84. But p v = 

etc.,= 84; hence, = —, /, = —, etc. When z/ = 1, 
84 

/ = ^=84=G^. 

When z»=2,/ = ^ = 43 = NB. 

/i 

84 

When z/=3,/ = ^ = 38 = 
o 

When z; = 4, / = ^ = 21 = Z Z>. 

4 

84 

When v=5,p = ~=z 16.8 = KE. 
o 

84 

When V = 6^ p = — = 14: = G F. 

A curve through Ay By Cy Dy Ey and F will be the one 
required. 


(570) Divide the space Q G into 10 equal parts, as shown 
in the figure, and draw the ordinate at the middle of each 
space. The average ordinate will be found to be 29.95 lb. The 
volume represented by the length Q G is 5 cu. ft. Hence, 
the work is 29.95 X 5 X 144 = 21,564 foot-pounds. Ans. 

Calculating the work by formula 79,, 

Z = 331.5744/F log-p; 


Z = 331.5744 X 84 X 1 X log j = 21,673 ft. -lb. Ans. 

(571) (a) 460 + 96 = 556° F. (d) 273^ -f 32 = 305C. 
(c) 273i + 180 = 453^ C. (d) 460 + 650 = 1,110° F. (e) 
273J - 40 = 233^° C. 


(572) 32°-20°=12°. 12 X 7 X .504 = 42.336 B. T. U. 
required to heat the ice to 32°. 142.65 X 7 =998.55 B. T. U. 

required to melt the ice. 212°— 32°= 180°. 180 X 7 Xl — 
1,260 B. T. U. required to heat the water to 212°. 
966.6 X 7= 6,766.2 B. T. U. required to change 7 lb. of 
water at 212° into steam at 212°. 42.336 + 998.55 + 1,260 + 


6 , 


766.2 = 9,067.086 B. T. 
9,067.086 X 778 
43 X 33,000 


U. 

= 4.971 horsepower. 


Ans. 



]74 


HEAT. 


(573) Using formula 84, and substituting the values 
of p and V at beginning of compression, we have 

A = 351.36/F 1^1] = 

351.36 X 14.7 X J- 

Log log 4-" = .41 log 4 = .41 X .60206 = .24684. 

The number whose logarithm is .24684 is 1.7654. 

Therefore, (-|gy“= 1.7654. 1 - (^g) = 1- 1-7654 = 

— .7654, the minus sign indicating that the air is compressed. 
Hence, Z = 351.36 X 14.7 X .7654 = 3,953.28 ft.-lb. Ans. 

(574) On account of the great irregularity in cutline 
of this figure, a division into ten parts will not give a suffi¬ 
ciently close approximation to the mean ordinate; hence, it 
is divided into 20 equal parts, as shown by the full lines. 



(See Fig. 47.) The dotted lines, situated midway between 
the full lines, represent the ordinates which are to be meas¬ 
ured. The sum of all the dotted lines, divided by the num¬ 
ber of divisions, gives .9365" for the mean ordinate. Draw 








































HEAT. 


175 


the line C D parallel to ^ B at 3. distance from A B equal 
to .9365'', and where it cuts the curve will be the points 
from which to draw the mean ordinate, as C E. 

(575) Prolong A B in both directions as shown in 
Fig. 48. Draw the tangents /i / Ay e dy and c by perpen¬ 
dicular to A B. Since the outline f A g very nearly 
triangular, and is quite small compared with the rest of the 
figure, consider it a triangle, and draw in n half way 



between A and g. Consider in n as the mean ordinate. 
Then, A g y, m ;2 = .34 x.28=: .0952 sq. in. In a similar 
manner, area of a d e = .17 X A = .06S sq. in. Dividing 
A f e d into 8 equal parts, and drawing the ordinates at the 
middle points of these divisions (for convenience the full 
lines, similar to those in the last figure, have been omitted, 
and the ordinates at the middle points only have been 
drawn), the mean ordinate is found to be 1.228^ The 
length A = 1.31"; hence, area of^ / ^^= 1.31 X 1.228 = 
1. 6087 sq. in. Dividing a c b into 8 equal parts, the mean 





176 


HEAT. 


ordinate is found to be .154; the length a ^ is 1.34; hence, 
area ^ = 1.34 X .154 = .2064 sq. in. Dividing h gi into 

5 equal parts, the mean ordinate is found to be .82"; the 
length h ^ is 1.07"; hence, area h g i— 1.07 X .82 = .8774 
sq. in. Dividing hih c into 10 equal parts, the mean ordi¬ 
nate is found to be 1.925"; its length h ^=3.21"; hence, 
area 4 ^ ^ = 3.21 X 1.925 = 6.179 sq. in. 

1.G087 + 6.179 = 7.7877 sq in. = area hi b c d ef A k. 

.0952+ .068 + .2064+ .8774 = 1.247 sq. in. 

7.7877 — 1.247 = 6.5407 sq. in. = required area bae fg, 

Ans. 


(576) Using formula 73, t = 




.?,+ IV , Sg 

For water, s^ = 1; therefore, -^5+ ^3*^3) 

+ and 

' t-t. 


Substituting the given values for W and t and the values of 
5‘j, .jg, etc., from Table 21, we have 


W = 


4X.426X80+.5X.0939 X 73-75.61(4X.426+.5X.0939) 
75.61 - 73 


7.35802 

2.61 


= 2.819 lb. 


Ans. 


(577) See Arts. 1138, 1139, 1131, 1132, and 
1152. 




Wv^ 120X1,200’ 




2X32.16 


= kinetic energy in 


(578) 

foot-pounds. 

(See Art. 957.) " 2 + 32 16 ^ “ kinetic energy ex¬ 

pended in heat. Now, dividing by 778 to reduce the foot- 

, ^ ^120 X 1200’X .15 a 

poundstoB.T.U.,. ^^3^ ^g^^^g = 517.975 B.T.U. Ans. 

(579) See Art. 1126. 

(580) Using formula 73, t = 

(Since j,, the specific heat of water, is 1, it may be left out.) 











HEAT. 


177 


Transposing, 

_ - /) + ITs J 3 (/s - f) _ 1.875(91-86)+1.25x.0562x(91-86) 

5(86-40) 

.0423. Ans. 

(See example, Art. 1137.) 


(581 ) r, = 460° + 450° = 910°; 7, = 460° + 70° = 530' 


Efficiency = 


T. - t : 


910 - 530 


= 41.76^. Ans. 


r, 910 

(582) Use formula 71,/ F= T". 7 = 460° + ^00 

= 660°; and A, from Table 20, is 5.34946; therefore, W =■ 
pV _ 20 X 700 

R T ~ 5.34946 X 660 


= 3.9653 lb. Ans. 


(583) (See Arts. 1091 to 1098.) 

{a) C ■.R = 100-.%0,ot C=^R=^^R. 

Hence, 44 X t = 55" C. Ans. 

4 

(b) F:R = n0-.m,oxF=^R = ^^R. 

Hence, 44 X ^ + 33 = 131° F. Ans. 
4 


(584) By formula 81,/, =/, The volume 

of 1 lb. of air at atmospheric pressure, and having a tem¬ 
perature of 60°, is 0^0296 ~ ques¬ 

tion 510.) 

Substituting in the above formula, 14.7 X 13.107‘*** = 
235 X or 


1.41 r 


14.7 X 13.107* 


235 


= 1.8356 cu. ft. Ans. 


From formula 71, remembering that 1F= 1, we have / V 
= A r, or T = ^-^= = 1,164-2°. 1,164.3°- 


460° = 704.2°. 


R 

Ans. 


(58S) (See Arts. 1107, 1054, and 1099.) 











178 


HEAT. 


(586) {a) Cubical contents of cylinder = (10’ — 9J’) 

816 48 

X .7854 X 72 = 816.48 cu. in. = -p^^u. ft. Specific grav¬ 
ity of copper is 8.79. Then, the weight of the cylinder is 

5^^^X 62.5 X 8.79 = 259.58 lb. Specific heat of copper, 
1,728 

from Table 21, is .0951. Substituting the values of n, W, 
and J, in formula 72, ^ — ^) we have 7,000 = 

259.58 (t, - /) X .0951, or / - / = 
increase of temperature. 

By formula 67, I = L C,t = nx .00000955 X 283.56 = 
.195 in. Ans. 

{b) By formula 69, v=V / = 816.48 X .00002864 X 
283.56 = 6.63 cu. in. Ans. 

(c) By formula 67, / = Z CT, /= 10 X .00000955 X 283.56 
= .027 in. Ans. 

(587) From Table 19, the coefficient of expansion for 
gases is .00203252. Substituting in formula 70, 

ri + C, (Z - 32)-l ^ ri + .00203252(390 - 32)1 

^ Li + Z3(^,-32^ ’ L 1 + .00203252(65 - 32) J ^ 

12 = 19.428 cu. ft. 19.428 - 12 = 7.428 cu. ft. Ans. 

The same answer may be obtained by using formula 58. 


(588) Foot-pounds per minute = 75 X 2 X 20 X 2 X .18. 
Foot-pounds per hour = 75 X 2 X 2 X 20 X .18 X 60. 


Heat units per hour = 
B. T. U. Ans. 


75 X 2 X 2 X 20 X .18 X 60 
778 


83.29 


(589) (a) Using formula 87, 7',= = 610 X 

(^)'‘‘ = -130.76°. Ans. 


{b) Substituting in formula 71, the initial values of 

ixr 'T Air • >. WRT .5x.37052X610 

fr, A, 7; and F, we obtain p = — y — =- - -= 

125.565 lb. per sq. in., the initial pressure. Ans. 










HEAT. 


179 


(^) Using formula 86, 

( T X.Trirr /329.24\ 

A =/ (y-) = 125.565 X = 15.06 


06 lb. 
per sq. in. 
Ans. 

12® X .5236 = 904.78 cu. in. = volume of sphere. 
904.78 


(590) 

Sp. Gr. of zinc = 6.86; therefore, 
224.5 lb. = weight of sphere 


X 62.5 X 6.86 = 


1,728 

(See example 586.) Using 


Wmnta'7't . IF,^,/.+ lF,5,/, _ 8X212 + m.5X.0956X70 
formula IS, t - ^ ^ ^ ^ggg 

= 108.49°. 108.49°— 70°= 38.49°= increase in temperature 

of the zinc sphere after dipping in the boiling water. 

Using formula 69, v = V Cj — 904.78 X.00004903 X 
38.49 = 1.71 cu. in., nearly. Ans. 


(591) By formula 67, l — L t. 6^, = .00000599; 
A = 900 X 12; / = 90° - 28°; therefore, 1 = 900 X 12 
X .00000599 X (90 — 28) = 4.01 in. Ans. 


(592) According to formula 72, ^ = B^(/, —/).?, or 
^ ~ W{t^ - 

(593) See Art. 1161. 

(594) {a) Temperature of vaporization of sulphur, from 
Table 22, is 228.3°. 

Specific heat of sulphur, from Table 21, is .2026. 

Latent heat of fusion of sulphur, from Table 22, is 13.26. 

228.3 - 40 = 188.3. 13 X .2026 X 188.3 = 495.94454 

B. T. U. to raise sulphur to melting point. 13.26 X 13 
= 172.38 B. T. U. to melt the sulphur. (495.94454 
4-172.38) X 778 X 519,956.5 ft.-lb., total work required to 
perform both of the above operations. Ans. 

,,, 519,956.5 , , . 

10 X 33,000 = horsepower. Ans. 


(595) 124 X 3 = 372 B. T. U., heat given up by the 

turpentine in changing from gas to liquid. 75 X 4 = 300 
B. T. U., heat contained in the water. 300 + 372= 672 








180 


HEAT. 


B. T. U., heat contained in the mixture at the instant the 
turpentine has become liquid. 672 4 = 168° = tempera- 

ture of water at the instant that the turpentine has become 
a liquid. ^ ^ 

Applying formula 73, / = " 

.3 X .426 X 313 -f 168 X 4 _ 203.11°, final temperature. Ans. 

3 X .426 + 4 

(596) Specific heat of melted lead, from Table 21,= .0402. 

Specific heat of solid lead, from Table 21, = .0314. 

Temperature of fusion of lead, from Table 22, = 626°. 

Latent heat of fusion of lead, from Table 22, = 9.67. 

626° - 46° = 580°. 800° - 626° = 174°. 

25 X .0314 X 580 = 455.3 B. T. U., heat required to raise 
the lead to melting point. 9.67 x25 = 241.75 B. T. U., 
heat required to melt the lead. 25 X .0402 X 174= 174.87 
B. T. U., heat required to raise the temperature of the 
melted lead from fusion point to 800°. 455.3 + 241.75 

+ 174.87 = 871.92 B. T. U., total heat required. 871.92 
X 778 = 678,353.76 ft.-lb.' Ans. 

(597) 32°—10°=22°. 22 X 10 X .504 = 110.88 B. T. U., 
heat required to raise ice from 10° to 32°. Latent heat of 
fusion of ice, from Table 22, is 142.65; then 142.65 X 10 
= 1,426.5 B. T. U., heat units to melt the ice. 1,426.5 
+ 110.88 = 1,537.38 B. T. U., total heat units to be taken 
from the mixture to melt the ice. Hence, applying formula 
73, 

, IVi Si ti + s^ A + ^3 /3 4- W<1 /4 + /s - 1,537.38 

WiSi+ W^s^+ Wz+ W,+ W,s, - 

11.5 X .1298 X 180 + 42 X .0939 X 240 + 10 X 32 + 50 x 120 + 20 X .0314 X 80 - 1,687.38 
11.5 X .1298 + 42 X .0939 + 10 + 50 + 20 X .0314 

= 91.55®. Ans. 

140 V .3*'“ 

(598) (a) A V,' " = A A"", or A = -++- = 13.21& 

Area = ^ 140x3-13.215x16 ^ 

.41 .41 

508.69 ~ 20 = 25.434 sq. in. Ans. 









HEAT. 


181 


(^) 25.434 -^ 13 = 1.9565. Ans. 

(c) 1.9565 X 20 = 39.13 lb. per sq. in. Ans. 

(699) Use formula 66, 4 = ((r ” I"* 

{a) 4 = ( - 10 - 32) X I = - 23i° C. Ans. 

{b) 4 = (25 - 32) X I = - 3f ° C. Ans. 


(c) 4 = (2,200 - 32) X I = 1,204|° C. Ans. 
y 

(600) This must be answered by the student. 

(601) Using formula 80, A = 331.5744/log ^ = 


331.5744 X 15 X 10 X log = 37,467.74 ft.-lb. Ans. 
io 

(602) 520 H. P. = 520 X 33,000 ft.-lb. per minute = 

520 X 33,000 X 60 ft.-lb. per hour = 

520 X 33,000 X 60 


778 


B. T. U. = 1,323,393.31 B. T. U. Ans 


(603) Let / = the required latent heat; 

= given weight of zinc; 

= given weight of water; 

= specific heat of zinc; 

= specific heaF of water ( = 1); 
/j = temperature of melting zinc; 
= temperature of water. 


Then, the total heat of the mixture is W\ -f 

W^l and from formula 73, the temperature 
__ lV,sJ,+ lVj, + lV,b 
lV,s,+ iV, 

IV, 

102^(4 X .0956 -I- 10) - 4 X .0956 X 680 - 10 X 60 _ 

^ Ans. 


(604) See Arts. 1H 7 and 1120. 

(605) («) and {b) See Art. 1118. 






182 


HEAT. 


(c) A non-conductor is one which will not conduct heat. 
There really is no such substance, but some substances are 
such bad conductors that they are termed non-conductors. 

(606) When answering this question, consult Art. 1123. 

(607) In Art. 1130 it is stated that 1 calorie = 3.96 

B. T. U. Hence, (a) 798 B. T. U. = 798 ~ 3.96 = 201.515 

calories. Ans. 


(d) 40 X 3.96 = 158.4 B. T. U. Ans. 

(608) (a) First calculate the weight by formula 71. 

/F 18X7.68 . 

: 33552 X 500 = 


Units of heat required = s„ W — T)~ s„W—t) 
= .15507 X .824 X (416 - 40) = 48.0444 B. T. U. 

Therefore, 48.0444 X 778 = 37,379 ft.-lb., nearly. Ans. 

(b) .21751 X .824(416-40) X 778 = 52,429ft.-lb., nearly. 

Ans. 

(609) See Arts. 1144 to H 47. 

(610) See Arts. 115 5 and 1164. 

(611) Using formula 87, 

T^= = 528^5^^ = 839.64°= 380°, nearly, above 

0° F. AnL 


(612) (See Arts. 1149 and 1177.) 

(613) Use formula 81 ,/ 1 '*'" = /, 

14.7 X 48'-" =/, X (48 - 38)'-“, or 

( 4 0 \ 1.41 

—j = 134.24 lb. per sq. in. Ans. 

Now, using formula 86, 

( 1.34 24 \ 

— y J = 951°, nearly, = 491° above 0° F. 

’ ' Ans 




Gas, Gasoline, and Oil Engines. 

(QUESTIONS 614-669.) 


(614) See Art. 1233. 


(615 ) To find the thermal efficiency, it is necessary to 
first calculate the maximum temperature. Using the adia¬ 
batic formula for temperature and volume, 


T V 




_ 1,600 X 1 _ 1,600 


.07-^^ .3367 

/;= r, = 4,760°. 


= 4,760‘ 


Calling the final temperature 7"^, 

^3 - ^4 _ 4,760 - 530 _ 4,230 
~ T; “ 4,760 “4,760 


:8887. (Art. 1242.) 


Hence the efficiency is 88.87^. Ans. 


(616) See Art. 1230. 

(617) {a) According to formula 90, Art. 1199, K = 

4.76 ; hence, since n = 4 and w = 6, C\//^ requires 

4.76 (4-}- 1|^) cu. ft. = 26.18 cu. ft. of air to 1 of the gas. 
Then 7 cu. ft. require 

7 X 26.18 cu. ft. = 183.26 cu. ft. Ans. 

(d) Likewise, requires 4.76 (3 -|- 2) cu. ft. = 23.8 cu. ft. 
of air to 1 of the gas. 23.8 cu. ft. X 15 = 357 cu. ft. Ans. 

(c) requires 4.76 (5 + 3) cu. ft. = 8 X 4.76 cu. ft. = 

38.08 cu. ft. of air to 1 of the gas. 38.08 cu. ft. x 77 = 
2,932.16 cu. ft. Ans. 

For notice of copyright, see page immediately following the title page. 






184 GAS, GASOLINE, AND OIL ENGINES. 

(618) First find the heat value of the gas from the 
performance of engine No. 1. Thus, gas consumption per 

jiour _ 17 ) cu. ft. per horsepower hour. 

18 


Then, according to formula 96, 

B — = 598.8. 

17 X .25 4.25 

The heat value of the gas is 598.8 B. T. U. 


The gas consumption of No, 2 = cu. ft. = 18 cu. ft. 
per horsepower hour. 

Then, by formula 94, the duty of engine No. 2 is 


D 


2,545 


2,545 


.2361+ , or 23.61+ i. Ans. 


598.8 X 18 10,778.4 

This example may also be solved by proportion thus: 

Let = duty engine No. 1; 

— duty engine No. 2. 

Then, 18 : 17. 


Hence, D- 


X 17 _ .25 X 17 
18 “ 18 


.2361, as before. 


(619) See Art. 1210. 

(620) (^) The power applied to the piston of the engine 
due to the expansion of the gases. 

(/;) The power delivered by the engine to the belts or 
other driving gear. (Art. 1246.) 

(621) In the diagram. Fig. 49, the difterent properties 
of an explosion .in a closed vessel are shown. 

The ignition occurs at a ; the time elapsing during the 
rise of pressure, or from a to is the duration of the explo¬ 
sion. From b to c the pressure is nearly constant, and the 
time elapsing from b to c \?> called the duration of maximum 
pressure. The time elapsing during the fall of pressure 







GAS, GASOLINE, AND OIL ENGINES. 185 


from ^ to <7^ is the duration of fall of pressure. The rapidity 
with which the pressure rises is known as the rate of flame 



propagation, and the rapidity with which the pressure falls 
is called the rate of fall of pressure. Both these rates are 
shown by the slopes of the lines. 


QQ K 

(622) A* = ^ = . 7857, or 78.57^. Ans. (Art. 1248.) 

(623) Total efficiency = .45 X .72 X .80 = .2592 = 
25.92^. Ans. (Art. 1241.) 

(624) Acetylene has a heat value of 868 B. T. U. Then, 
according to formula 94, 


A = 


2,545 
868 X 17 


2,545 

14,756 


= .1724. 


Hence the duty is 17.24^. Ans. 

(626) (a) See Art. 1226. 

(d) See Arts. 1228 and 1229. 

(626) (^) See Art. 1205. 

(5) The expansion line of an actual gas-engine diagram ‘ 
follows the theoretical line much more closely than if disso¬ 
ciation was not present. (Art. 1236.) 

(627) {^) The horsepower is proportional to the num¬ 
ber of the heat units in the gas. 

Let X = the power when using Pittsburg gas. 

Then, x : 10 :: 892 : 800, 


11.15 H.P. Ans. 

800 


U.G. IV.—12 






18G GAS, GASOLINE, AND OIL ENGINES. 


(/>») Let y = the horsepower when using lo-candle-power 
gas. 

Then j : 10 :: 620 : 80, 


y = 


620 X 10 
800 


7.75 H. P. 


Ans. 


(628) (a) See Art. 1243. 

(^) Duty and thermal efficiency are practically the same. 

(c) The ratio of the D. H. P. to the I. H. P. 

(629) (a) Compression of a charge of gas and air, 
explosion and expansion of the same. 

(d) Compression of a charge of pure air until a high 
pressure and temperature is obtained; admission of fuel 
which is ignited by the hot air and burns as it enters; 
expansion of the products of combustion. 

187 5 

(630) Gas consumption per horsepower hour = ■ —- ' - = 

1/C. 0 

O KA^ 

15 cu. ft. By formula 94, D = —— = .2121, or 21.21^. 

lo X oUU 

(Art. 1245.) Heat value of Leechburg gas = 1,051 B.T.U. 
Then, by formula 95, 


G 


2,545 

.2121 X 1,051 


cu. ft. = 11.41-|- cu. ft., 


the gas consumption per H. P. 


11.41 cu. ft. X 15 = 171.15 cu. ft. Ans. 


(631) The governor controls the gas supply by shifting 
the cam actuating the gas-valve. If the speed gets too high, 
the cam is moved so that it will not operate the gas-valve 
lever, and the gas-valve remains closed until the speed falls. 

(632) The amount of work stored in the oil is found by 
multiplying the number of pounds used per day by the heat 
value per pound and by Joule’s equivalent, 778. Thus, 
4i X 6.8 X 17,933 X 778 = 426,927,344.4, the number of foot¬ 
pounds stored. Work delivered per minute = 33,000 H. P. = 






GAS, GASOLINE, AND OIL ENGINES. 


187 


33,000 ft.-lb. X 3 = 90,000 ft.-lb. ; the work delivered per 
day of 10 hours is therefore 99,000 ft.-lb. X 10 X 60 = 

59,400,000 ft.-lb. Total efficiency = 344 4 = 

594 

.1391, or 13.91^. Ans. 


(633) Because of the rapid escape of heat through the 
walls of the cylinder. 

(634) {a) See Fig. 243, Art. 1232. 

(d) See Fig. 244, Art. 1234. 

(635) A compound that explodes by instantaneous 
chemical combination, and that can be exploded by a jar. 


(636) (ci) Call the duty of each engine 
respectively. Then, 

Engine No. 1 consumes 185 ^ 10 = 18.5 cu. ft. per H. P. 
hour. Then, using formula 94, / 

2 545 

^■ = 18.5X 690 = -^^^^’ 

Engine No. 2 consumes 240 15 = 16 cu. ft. per H. P. 

hour. 

o K4.K 


Engine No. 
hour. 

A- 


3 consumes 540 -h 45 = 12 cu. ft. per H. P. 
2 545 


(b) The engine having the highest duty, i. e., engine 
No. 3. 

(637) (a) Applying the law for pressure and volume 

P F'-*" 

with adiabatic compression, —, 


or 


F = 



14.7 X 
185 


Ans. 










188 GAS, GASOLINE, AND OIL ENGINES. 


Using logarithms, we have 

log 14.7 + 1.41 log4 — log 185 .25095 

log h, - - “2 41 

- .17798 = 1.82202. 

Hence, = .6638 cu. ft. 

(,) = = 1.4.37“ absolute. 


F.-‘ 


Ans. 


(638) (a) The union of two or more substances to form 
a chemical compound. 

(^) The combination of CH^ with (9, forming CO^ and 
Hfi is a good example. 

(639) {a) The chemical combination of two or more 
substances with the production of light and heat. 

{b) Compounds formed by combustion. 

(c) Carbon dioxide {CO^), water etc. 

(640) See Arts. 1262 and 1263. 

(641) 

P V 1 4 7 V 1 

(a) — — '-^,. 41 - ■ = 142.2 lb. absolute. Ans. 



r.= 

Tyr 

530 X 1 


y," 

.2-** 

(i) 

T — 

T,P,_ 

1,025 X 375 


p. ~ 

142.2 . 


7 ’.= 


2,703 X .2- 

vr 

1 


: 1,025-}-° absolute. Ans. 
= 2,703° absolute. Ans. 


Ans. 

P V ^7^ V 9*-^* 

P i — • —- Y~ -~ absolute. Ans. 

(rf) .4828, or 48.28^. 


Ans. 


(642) See Art. 1271. 

















GAS, GASOLINE, AND OIL ENGINES. 189 

(643) Applying formula 91, 

V= 1.52 (C+dN), 

F = 1.52 (57 + 126) = 1.52 (183) = 278.16, the number of 
cu. ft. of air to 1 lb. of oil. 

278.16 cu. ft. X 27 = 7,510.32 cu. ft. Ans. 

(644) (a) The contents of the cylinder before ignition. 

(d) The gases expelled from the cylinder. 

(c) That part of the cylinder back of the piston into 
which the piston does not enter. 

(645) (a) See Art. 1235. 

(3) It is much higher. 

(646) {a) See Art. 1189. 

(3) See Art. 1189. 

(<:) Hydrogen and sulphur are elements; hydrogen sul¬ 
phide and carbon dioxide are compounds. 

(647) Referring to Table 23, the amount of air required 
for the complete combustion of the gas is computed as 
follows : 

H, .061 X 2.38 = .14518 

.7544 X 9.52 = 7.18188 
.0935 X 14.28 = 1.3352 
.0877 X 23.82 = 2.0890 

10.75126 

Call the amount 10.751 cu. ft. 

With 10^ excess, the quantity required is 10.751 cu. ft. X 
1.10 = 11.826 cu. ft. Ans. 

(648) See Art. 1213. 

(649) By shifting the exhaust cam, either by hand or 
by the aid of a lever called a starting lever. The exhaust 
valve is then automatically held open until half the com¬ 
pression stroke is completed. 

(650) See Art. 1212. 



190 GAS, GASOLINE, AND OIL ENGINES. 

(651) Applying formula 90 for hydrocarbons, we have: 

(<r) C,//, requires 4.70 (3 + 1) cu. ft. = 14.28 cu. ft. of 

air to 1 of gas; hence,the volume required is 14.38 cu. ft. X 
25 = 357 cu. ft. Ans. 

(b) requires 4.76 (6 + 1^) cu. ft. = 35.7 cu. ft. of air 
to 1 of gas. 35.7 cu. ft. X 7 = 249.9 cu. ft. Ans. 

(c) requires 4.76 (11 + b^) cu. ft. = 78.54 cu. ft. of 
air to 1 of gas. 78.54 cu. ft. X 65 = 5,105.1 cu. ft. Ans. 

(652) Applying formula 92, 

we have .08 X 293.5 = 23.48 

.1633 X 868 = 141.7444 

CO, .695 X 674 = 468.43 

Gasoline, .0617 X 690 = 42.573 

676.2274 B.T.U. Ans. 

(653) Carbon dioxide (CO^), water (H^O), and sulphur 
dioxide (vSt^J. 

(654) (a) The number of B. T. U. given up by the 
combustion of 1 cubic foot of the gas. 

(I?) The British thermal unit, B. T. U. 

(655) (a) Four. 

(A Type III. 

(c) Type IV. 

(656) The modern engine has poppet-valves instead of 
the slide-valve, and employs electric ignition instead of 
flame ignition*. 

(657) See Arts. 1219 to 1221. 

(658) (^) The power required to drive the engine itself. 

(d) By running the engine without load and finding the 
indicated horsepower. By subtracting the D. H. P. from 
the I. H. P. 



GAS, GASOLINE, AND OIL ENGINES. 191 

(659) («) The space surrounding the cylinder, for cir¬ 
culation of cooling water. 

{b) To prevent overheating of the cylinder and piston. 

(660) (^) An engine requiring four strokes to complete 
the cycle. 

{b) By means of bevel-gearing or worm-gearing. 

(661) {a) and {b) See Art. 1206. 


(c) See Art. 1209. 

(662) See Art. 1252. 

(663) (^) Applying the adiabatic formulas we have 



8^ of the stroke = 93.5 X .08 = 7.48^ of cylinder volume. 

7.48^ -f 6.5^ = 13.98^ = = volume behind the piston at 

cut-off. 

Next applying formula 53, 


P,V, _ 693.6 X .065 


F, .1398 


= 322.5 lb. 



T,= 7 ; = 1,610°. 

322.5 X .1398'*'" , 


= 20.13 lb. absolute. Ans. 


1,610 X. 1398" 


= 718.70°, nearly, absolute. Ans. 



r, _ 1,610-718.7 
^3 “ 1,610 


= .5536, nearly. Ans. 


(664) See Arts. 1255 and 1256. 

(665) vSee Art. 1266. 











m GAS, GASOLINE, AND OIL ENGINES. 

(666) See Art. 1257. 

(667) (a) See Art. 1268. 

(6) See Art. 1269. 

(668) SeeArt. 1261. 

(669) See Arts. 1258 and 1259. 


Gas, Gasoline, and Oil Engines. 

(QUESTIONS 670-709.) 


(670) (a) See Arts. 1370 and 1371. 

{b) See Art. 1372. 

(671) By the planimeter the area of the diagram is 
found to be 2 sq. in. The length is 3.22 in. Using 
formula 114, 

TVT t:' -n ^ X 1^0 • 1 A 

M. E. P. = -y- = — = 62 lb. per sq. m. nearly. Ans. 

The method of measurement described in Art. 1356 
may be used if a planimeter is not available. The result 
obtained may be slightly different from that here given. 

(672) See Art. 1292. 

(673) Since there are 4 cylinders, the H. P. developed in 
each is — 14. Using formula lOO to find the speed of 
the engine, we have 

_ ^ ^ 

- yy... - I4..r 

Using logarithms, 

log R - log 350 - . 21 log 14 = 2.30338, 

A= 201+, say, 200 R. P. M. 

From formula 99, 

ry^T _ 28,000 





194 GAS, GASOLINE, AND OIL ENGINES. 


Substituting for A, its value D, or we have 
28,000//' 




R 


or 


Ans. 


n _ iy28,000 X i X // _ //28,000 X i X 14 _ 

^^ R 200 

L = ^ D = ^ X 11.6 in. = 14.5 in. Ans. 

(674) See Arts. 1288, 1334, and 1335. 

(675) Using formula 113, 

■Pk xj D ^ r^p In 2x3.1416 X 25 X 5 X 180 _ ^ ^ 

^ 33,OW =-“ 

(676) (a) See Art. 1369. 

(b) See Art. 1373. 

(r) See Arts. 1370 and 1374. 

(677) The piston area is (.7854 X 14’) sq. in. = 
153.94 sq. in. Since the piston speed is 550 ft. per min., the 

mean velocity of the crank-pin is (—7 X 0 ft. per sec. 

\ 60 2 / 

Substituting the various values in formula 1 lO, 

A Hg _ 153.94 X 115 X 32.16 


W-. 


n^BV: 




= 4,068 lb. Ans. 


(678) Using formulas 102 and 103, 

A = .346 jD = .346 X 15 in. = 5.19 in., say 5-^ in. Ans. 

/' = .4 // = .4 X 15 in. = 6 in. Ans. 

According to formula 104, the lift in each case is of 
the diameter of the port; hence, 

lift of air-valve = 5y\ in. X i = lyV nearly. Ans. 


lift of exhaust-valve = 6 in. X J = 1J in. Ans. 

( 679 ) Substituting the given values in formula 114 , 

1-'^^ X 120 ^ 

M. h>. r. = =- - -= 59 lb. per sq. in., nearly. Ans. 


3.5 











GAS, GASOLINE, AND OIL ENGINES. 


195 


(680) Diagram (r?) shows too early ignition and too 
late opening of the exhaust valve; diagram (d) shows too 
late ignition; diagram (^') is practically faultless. 


(681) Using formula 115, 


Plan _6SX if X .7854 X 12^ X 90 
33,000“ 33,000 


= 27.97. Ans. 


(682) See Art. 1303. 

(683) Using formula 116, Art. 1361, 

H=Swq(T-t) =.225X.062X55X (425-74) = 269.3 B.T.U. 

Ans. 

(684) See Art. 1 290. 

(685) (a) See Art. 1279. 

{^) See Art. 1 285. 

(r) See Art. 1 282. 

( 686 ) Tangential force = 4,200 lb. X sin 35° = 4,200 lb. 
X .57358 = 2,409+ lb. Ans. 

Radial force = 4,200 lb. x cos 35° = 4,2001b. X .81915 = 
3,440.41b. Ans. 


(687) (a) See Art. 1294. 

(d) See Art. 1 298. 

{c) See Art. 1299. 

(688) (a) and (I?) See Art. 1316. 

{c) See Art. 1321. 

(689) Using formula 107, 

M= I— (3.5X 12’'X 15) cu. in.=7,560cu. in. = 4fcu. ft. 

Ans. 

(690) See Arts 1306, 1308, 1309, and 1310. 

(691) {a) Using formula lOl, 

„ _ 405 _ 405 
A - ^ 

Log R = log 405 — .21 log 35 = 2.2832; 
whence, R = 192. Ans. 




inC GAS, GASOLINE, AND OIL ENGINES. 

,, LR 

{b) Substituting in formula 98, tl — 
transposing, 


14,000’ 


and 




14,000 7/_ 14,000 X 35 


R 


192 


For L substitute \\D^ ox ^D\ then, 

14,000 X 35 


4 773 
3 ^ 


192 


and 


= = 12.42 in. Ans. 

Xi/zV 


Z = f = f X 12.42 = 16.56 in. Ans. 

In practice the engine would be made 12^' X 16^'. 

(692) Area of piston = .7854 X 15^^ = 176.72 sq. in. 
Area of inlet port = 12^ of piston area = 176.72 sq. in. X 
.12 — 21.2 sq. in. 

Using formula 105, 

, Zax 3 X 21.2 X 1 




= 6.36. 


2(j:+/) 2(4+1) 

Hence the area of the gas passages should be 6.36 sq. in. Ans. 


(693) See Art. 1354. 

(694) {a) See Art. 1376. 

(b) More gasoline should be turned on. 

(c) The needle valve should be partly closed. 

(695) See Art. 1333. 

(696) Applying formula 98, 


D^LR _ 10" X 14 X 200 
14,000 ~ 14,000 


= 20 . 


Ans. 


(697) {a) The oil is injected in a fine stream into a 

hot chamber. See Art. 1 388. 

{b) Ignition is caused by the contact with the hot walls 
of the vaporizer. 












GAS, GASOLINE, AND OIL ENGINES. 197 


(698) (a) See Art. 1291. 

{b) See Art. 1292. 

(699) Referring to the notation of Art. 1322, 

IF, = 4 + f X 40 = 29; 

W, = ZQ- 

IF, = 45 + 12 + 30 + f X 40= 102; 
h =3.9; 
r =6; 

R = 26. 

Substituting these values in formula 109, 



IF = 


= 23 lb. nearly. Ans. 


(700) The gasoline engine may be applied to the fol¬ 
lowing: Motor carriages, gasoline launches, mine hoists, 
traction engines. 

(701) See Art. 1321. 

(702) See Art. 1374. 

(703) See Art. 1322. 

(704) See Art. 1277. 

(705) In the method of vaporizing the oil, and in the 
method of igniting the charge. 

(706) See Art. 1356. 

(707) See Art. 1278. 

(708) See Art. 1336. 

(709) See Art. 1403. 









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